Free Printable Conservation of Mass Worksheet Collection - Free Printable
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Step-by-step solution for: Free Printable Conservation of Mass Worksheet Collection
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Step-by-step solution for: Free Printable Conservation of Mass Worksheet Collection
To balance the chemical equations according to the law of conservation of mass, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Here are the balanced equations for each problem:
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This reaction cannot occur as written because it violates the law of conservation of mass. Oxygen molecules (\( \text{O}_2 \)) cannot directly form ozone (\( \text{O}_3 \)) without additional reactants or intermediates. Ozone formation typically involves a three-step process involving free radicals.
Balanced Equation: Not possible as written.
---
- Start with 1 Zn and 1 HCl.
- On the right side, there are 2 Cl in \( \text{ZnCl}_2 \), so we need 2 HCl.
- Balance H: 2 HCl gives 2 H, which matches 2 H in \( \text{H}_2 \).
Balanced Equation: \( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \)
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- Start with 2 N in \( \text{N}_2 \).
- Each \( \text{NH}_3 \) has 1 N, so we need 2 \( \text{NH}_3 \).
- Balance H: 2 \( \text{NH}_3 \) have 6 H, so we need 3 \( \text{H}_2 \).
Balanced Equation: \( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \)
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- Start with 2 Cr in \( \text{Cr}_2\text{O}_3 \).
- Each \( \text{Cr} \) requires 1 Al, so we need 2 Al.
- Balance Al: 2 Al produces 1 \( \text{Al}_2\text{O}_3 \).
- Balance O: \( \text{Cr}_2\text{O}_3 \) has 3 O, which matches 3 O in \( \text{Al}_2\text{O}_3 \).
Balanced Equation: \( 2\text{Al} + \text{Cr}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + 2\text{Cr} \)
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- Start with 3 O in \( \text{KClO}_3 \).
- \( \text{O}_2 \) has 2 O, so we need \( \frac{3}{2} \) \( \text{O}_2 \). Multiply everything by 2 to eliminate the fraction.
- Balance K and Cl: 2 \( \text{KClO}_3 \) gives 2 K and 2 Cl.
Balanced Equation: \( 2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2 \)
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- Start with 1 B in \( \text{BF}_3 \).
- Each \( \text{B}_2\text{O}_3 \) has 2 B, so we need 2 \( \text{BF}_3 \).
- Balance F: 2 \( \text{BF}_3 \) have 6 F, so we need 6 HF.
- Balance H and O: 6 HF require 6 H, and 3 \( \text{H}_2\text{O} \) provide 6 H and 3 O for \( \text{B}_2\text{O}_3 \).
Balanced Equation: \( 2\text{BF}_3 + 3\text{H}_2\text{O} \rightarrow \text{B}_2\text{O}_3 + 6\text{HF} \)
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- Start with 1 P in \( \text{PCl}_5 \).
- Each \( \text{PF}_5 \) has 1 P, so we need 1 \( \text{PCl}_5 \).
- Balance As: 1 As in \( \text{AsF}_3 \) matches 1 As in \( \text{AsCl}_3 \).
- Balance Cl: \( \text{PCl}_5 \) has 5 Cl, which matches 3 Cl in \( \text{AsCl}_3 \) and 2 Cl in \( \text{PF}_5 \).
- Balance F: \( \text{AsF}_3 \) has 3 F, which matches 5 F in \( \text{PF}_5 \).
Balanced Equation: \( \text{PCl}_5 + \text{AsF}_3 \rightarrow \text{PF}_5 + \text{AsCl}_3 \)
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- Start with 1 K.
- Each \( \text{K}_2\text{O}_2 \) has 2 K, so we need 2 K.
- Balance O: \( \text{O}_2 \) has 2 O, which matches 2 O in \( \text{K}_2\text{O}_2 \).
Balanced Equation: \( 2\text{K} + \text{O}_2 \rightarrow \text{K}_2\text{O}_2 \)
---
- Start with 1 Fe.
- Each \( \text{Fe(OH)}_2 \) has 1 Fe, so we need 1 Fe.
- Balance O: \( \text{Fe(OH)}_2 \) has 2 O, which comes from 1 \( \text{O}_2 \) and 1 \( \text{H}_2\text{O} \).
- Balance H: \( \text{Fe(OH)}_2 \) has 2 H, which comes from 1 \( \text{H}_2\text{O} \).
Balanced Equation: \( \text{Fe} + \text{O}_2 + \text{H}_2\text{O} \rightarrow \text{Fe(OH)}_2 \)
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- Start with 1 Al.
- Each \( \text{AlCl}_3 \) has 1 Al, so we need 1 Al.
- Balance Cl: \( \text{AlCl}_3 \) has 3 Cl, so we need 3 HCl.
- Balance H: 3 HCl gives 3 H, which forms 1.5 \( \text{H}_2 \). Multiply everything by 2 to eliminate the fraction.
Balanced Equation: \( 2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2 \)
---
- Start with 1 N in \( \text{NH}_3 \).
- Each \( \text{N}_2\text{O}_3 \) has 2 N, so we need 2 \( \text{NH}_3 \).
- Balance H: 2 \( \text{NH}_3 \) have 6 H, so we need 3 \( \text{H}_2\text{O} \).
- Balance O: \( \text{N}_2\text{O}_3 \) has 3 O, and 3 \( \text{H}_2\text{O} \) have 3 O, so we need 3 O from \( \text{O}_2 \).
Balanced Equation: \( 2\text{NH}_3 + 3\text{O}_2 \rightarrow \text{N}_2\text{O}_3 + 3\text{H}_2\text{O} \)
---
- Start with 2 K in \( \text{K}_2\text{O}_2 \).
- Each \( \text{KOH} \) has 1 K, so we need 2 \( \text{KOH} \).
- Balance O: \( \text{K}_2\text{O}_2 \) has 2 O, and 2 \( \text{KOH} \) have 2 O, so we need 1 \( \text{O}_2 \).
- Balance H: 2 \( \text{KOH} \) require 2 H, so we need 1 \( \text{H}_2\text{O} \).
Balanced Equation: \( \text{K}_2\text{O}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{KOH} + \text{H}_2 + \text{O}_2 \)
---
- Start with 1 Mn.
- Each \( \text{MnCl}_2 \) has 1 Mn, so we need 1 Mn.
- Balance Cu: 1 Cu in \( \text{CuCl} \) matches 1 Cu on the right.
- Balance Cl: \( \text{CuCl} \) has 1 Cl, so we need 2 \( \text{CuCl} \) to get 2 Cl for \( \text{MnCl}_2 \).
Balanced Equation: \( \text{Mn} + 2\text{CuCl} \rightarrow 2\text{Cu} + \text{MnCl}_2 \)
---
- Start with 1 Mg in \( \text{Mg(OH)}_2 \).
- Each \( \text{Mg}_3(\text{PO}_4)_2 \) has 3 Mg, so we need 3 \( \text{Mg(OH)}_2 \).
- Balance P: \( \text{H}_3\text{PO}_4 \) has 1 P, so we need 2 \( \text{H}_3\text{PO}_4 \) to get 2 P for \( \text{Mg}_3(\text{PO}_4)_2 \).
- Balance O: 3 \( \text{Mg(OH)}_2 \) have 6 O, and 2 \( \text{H}_3\text{PO}_4 \) have 8 O, so we need 6 \( \text{H}_2\text{O} \) to balance the O.
- Balance H: 3 \( \text{Mg(OH)}_2 \) have 6 H, and 2 \( \text{H}_3\text{PO}_4 \) have 6 H, which matches 6 H in 6 \( \text{H}_2\text{O} \).
Balanced Equation: \( 3\text{Mg(OH)}_2 + 2\text{H}_3\text{PO}_4 \rightarrow 6\text{H}_2\text{O} + \text{Mg}_3(\text{PO}_4)_2 \)
---
- Start with 1 Ba in \( \text{Ba(HCO}_3)_2 \).
- Each \( \text{BaCO}_3 \) has 1 Ba, so we need 1 Ba.
- Balance C: \( \text{Ba(HCO}_3)_2 \) has 2 C, so we need 2 \( \text{CO}_2 \).
- Balance O: \( \text{Ba(HCO}_3)_2 \) has 6 O in \( \text{CO}_3 \) groups, 2 O in \( \text{BaCO}_3 \), and 2 O in 2 \( \text{CO}_2 \), so we need 1 \( \text{H}_2\text{O} \).
- Balance H: \( \text{Ba(HCO}_3)_2 \) has 2 H, which matches 2 H in \( \text{H}_2\text{O} \).
Balanced Equation: \( \text{Ba(HCO}_3)_2 \rightarrow \text{BaCO}_3 + \text{H}_2\text{O} + 2\text{CO}_2 \)
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- Start with 1 Zn.
- Each \( \text{ZnS} \) has 1 Zn, so we need 1 Zn.
- Balance S: \( \text{S}_8 \) has 8 S, so we need 8 \( \text{ZnS} \).
- Balance Zn: 8 \( \text{ZnS} \) require 8 Zn.
Balanced Equation: \( 8\text{Zn} + \text{S}_8 \rightarrow 8\text{ZnS} \)
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- Start with 1 Li in \( \text{LiHCO}_3 \).
- Each \( \text{Li}_2\text{CO}_3 \) has 2 Li, so we need 2 \( \text{LiHCO}_3 \).
- Balance C: 2 \( \text{LiHCO}_3 \) have 2 C, so we need 1 \( \text{CO}_2 \) and 1 \( \text{Li}_2\text{CO}_3 \).
- Balance O: 2 \( \text{LiHCO}_3 \) have 6 O, and \( \text{Li}_2\text{CO}_3 \) has 3 O, \( \text{CO}_2 \) has 2 O, so we need 1 \( \text{H}_2\text{O} \).
- Balance H: 2 \( \text{LiHCO}_3 \) have 2 H, which matches 2 H in \( \text{H}_2\text{O} \).
Balanced Equation: \( 2\text{LiHCO}_3 \rightarrow \text{Li}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \)
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- Start with 2 N in \( \text{N}_2 \).
- Each \( \text{N}_2\text{O}_5 \) has 2 N, so we need 1 \( \text{N}_2 \).
- Balance O: \( \text{N}_2\text{O}_5 \) has 5 O, so we need \( \frac{5}{2} \) \( \text{O}_2 \). Multiply everything by 2 to eliminate the fraction.
Balanced Equation: \( 2\text{N}_2 + 5\text{O}_2 \rightarrow 2\text{N}_2\text{O}_5 \)
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- Start with 5 C in \( \text{C}_5\text{H}_{12} \).
- Each \( \text{CO}_2 \) has 1 C, so we need 5 \( \text{CO}_2 \).
- Balance H: \( \text{C}_5\text{H}_{12} \) has 12 H, so we need 6 \( \text{H}_2\text{O} \).
- Balance O: 5 \( \text{CO}_2 \) have 10 O, and 6 \( \text{H}_2\text{O} \) have 6 O, so we need 8 \( \text{O}_2 \).
Balanced Equation: \( \text{C}_5\text{H}_{12} + 8\text{O}_2 \rightarrow 6\text{H}_2\text{O} + 5\text{CO}_2 \)
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- Start with 4 C in \( \text{C}_4\text{H}_8\text{O}_2 \).
- Each \( \text{CO}_2 \) has 1 C, so we need 4 \( \text{CO}_2 \).
- Balance H: \( \text{C}_4\text{H}_8\text{O}_2 \) has 8 H, so we need 4 \( \text{H}_2\text{O} \).
- Balance O: \( \text{C}_4\text{H}_8\text{O}_2 \) has 2 O, 4 \( \text{CO}_2 \) have 8 O, and 4 \( \text{H}_2\text{O} \) have 4 O, so we need 5 \( \text{O}_2 \).
Balanced Equation: \( \text{C}_4\text{H}_8\text{O}_2 + 5\text{O}_2 \rightarrow 4\text{H}_2\text{O} + 4\text{CO}_2 \)
---
\[
\boxed{
\begin{aligned}
1. & \quad \text{Not possible as written.} \\
2. & \quad \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \\
3. & \quad \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \\
4. & \quad 2\text{Al} + \text{Cr}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + 2\text{Cr} \\
5. & \quad 2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2 \\
6. & \quad 2\text{BF}_3 + 3\text{H}_2\text{O} \rightarrow \text{B}_2\text{O}_3 + 6\text{HF} \\
7. & \quad \text{PCl}_5 + \text{AsF}_3 \rightarrow \text{PF}_5 + \text{AsCl}_3 \\
8. & \quad 2\text{K} + \text{O}_2 \rightarrow \text{K}_2\text{O}_2 \\
9. & \quad \text{Fe} + \text{O}_2 + \text{H}_2\text{O} \rightarrow \text{Fe(OH)}_2 \\
10. & \quad 2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2 \\
11. & \quad 2\text{NH}_3 + 3\text{O}_2 \rightarrow \text{N}_2\text{O}_3 + 3\text{H}_2\text{O} \\
12. & \quad \text{K}_2\text{O}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{KOH} + \text{H}_2 + \text{O}_2 \\
13. & \quad \text{Mn} + 2\text{CuCl} \rightarrow 2\text{Cu} + \text{MnCl}_2 \\
14. & \quad 3\text{Mg(OH)}_2 + 2\text{H}_3\text{PO}_4 \rightarrow 6\text{H}_2\text{O} + \text{Mg}_3(\text{PO}_4)_2 \\
15. & \quad \text{Ba(HCO}_3)_2 \rightarrow \text{BaCO}_3 + \text{H}_2\text{O} + 2\text{CO}_2 \\
16. & \quad 8\text{Zn} + \text{S}_8 \rightarrow 8\text{ZnS} \\
17. & \quad 2\text{LiHCO}_3 \rightarrow \text{Li}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \\
18. & \quad 2\text{N}_2 + 5\text{O}_2 \rightarrow 2\text{N}_2\text{O}_5 \\
19. & \quad \text{C}_5\text{H}_{12} + 8\text{O}_2 \rightarrow 6\text{H}_2\text{O} + 5\text{CO}_2 \\
20. & \quad \text{C}_4\text{H}_8\text{O}_2 + 5\text{O}_2 \rightarrow 4\text{H}_2\text{O} + 4\text{CO}_2 \\
\end{aligned}
}
\]
---
1. \( \text{O}_2 \rightarrow \text{O}_3 \)
This reaction cannot occur as written because it violates the law of conservation of mass. Oxygen molecules (\( \text{O}_2 \)) cannot directly form ozone (\( \text{O}_3 \)) without additional reactants or intermediates. Ozone formation typically involves a three-step process involving free radicals.
Balanced Equation: Not possible as written.
---
2. \( \text{Zn} + \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \)
- Start with 1 Zn and 1 HCl.
- On the right side, there are 2 Cl in \( \text{ZnCl}_2 \), so we need 2 HCl.
- Balance H: 2 HCl gives 2 H, which matches 2 H in \( \text{H}_2 \).
Balanced Equation: \( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \)
---
3. \( \text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3 \)
- Start with 2 N in \( \text{N}_2 \).
- Each \( \text{NH}_3 \) has 1 N, so we need 2 \( \text{NH}_3 \).
- Balance H: 2 \( \text{NH}_3 \) have 6 H, so we need 3 \( \text{H}_2 \).
Balanced Equation: \( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \)
---
4. \( \text{Al} + \text{Cr}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + \text{Cr} \)
- Start with 2 Cr in \( \text{Cr}_2\text{O}_3 \).
- Each \( \text{Cr} \) requires 1 Al, so we need 2 Al.
- Balance Al: 2 Al produces 1 \( \text{Al}_2\text{O}_3 \).
- Balance O: \( \text{Cr}_2\text{O}_3 \) has 3 O, which matches 3 O in \( \text{Al}_2\text{O}_3 \).
Balanced Equation: \( 2\text{Al} + \text{Cr}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + 2\text{Cr} \)
---
5. \( \text{KClO}_3 \rightarrow \text{KCl} + \text{O}_2 \)
- Start with 3 O in \( \text{KClO}_3 \).
- \( \text{O}_2 \) has 2 O, so we need \( \frac{3}{2} \) \( \text{O}_2 \). Multiply everything by 2 to eliminate the fraction.
- Balance K and Cl: 2 \( \text{KClO}_3 \) gives 2 K and 2 Cl.
Balanced Equation: \( 2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2 \)
---
6. \( \text{BF}_3 + \text{H}_2\text{O} \rightarrow \text{B}_2\text{O}_3 + \text{HF} \)
- Start with 1 B in \( \text{BF}_3 \).
- Each \( \text{B}_2\text{O}_3 \) has 2 B, so we need 2 \( \text{BF}_3 \).
- Balance F: 2 \( \text{BF}_3 \) have 6 F, so we need 6 HF.
- Balance H and O: 6 HF require 6 H, and 3 \( \text{H}_2\text{O} \) provide 6 H and 3 O for \( \text{B}_2\text{O}_3 \).
Balanced Equation: \( 2\text{BF}_3 + 3\text{H}_2\text{O} \rightarrow \text{B}_2\text{O}_3 + 6\text{HF} \)
---
7. \( \text{PCl}_5 + \text{AsF}_3 \rightarrow \text{PF}_5 + \text{AsCl}_3 \)
- Start with 1 P in \( \text{PCl}_5 \).
- Each \( \text{PF}_5 \) has 1 P, so we need 1 \( \text{PCl}_5 \).
- Balance As: 1 As in \( \text{AsF}_3 \) matches 1 As in \( \text{AsCl}_3 \).
- Balance Cl: \( \text{PCl}_5 \) has 5 Cl, which matches 3 Cl in \( \text{AsCl}_3 \) and 2 Cl in \( \text{PF}_5 \).
- Balance F: \( \text{AsF}_3 \) has 3 F, which matches 5 F in \( \text{PF}_5 \).
Balanced Equation: \( \text{PCl}_5 + \text{AsF}_3 \rightarrow \text{PF}_5 + \text{AsCl}_3 \)
---
8. \( \text{K} + \text{O}_2 \rightarrow \text{K}_2\text{O}_2 \)
- Start with 1 K.
- Each \( \text{K}_2\text{O}_2 \) has 2 K, so we need 2 K.
- Balance O: \( \text{O}_2 \) has 2 O, which matches 2 O in \( \text{K}_2\text{O}_2 \).
Balanced Equation: \( 2\text{K} + \text{O}_2 \rightarrow \text{K}_2\text{O}_2 \)
---
9. \( \text{Fe} + \text{O}_2 + \text{H}_2\text{O} \rightarrow \text{Fe(OH)}_2 \)
- Start with 1 Fe.
- Each \( \text{Fe(OH)}_2 \) has 1 Fe, so we need 1 Fe.
- Balance O: \( \text{Fe(OH)}_2 \) has 2 O, which comes from 1 \( \text{O}_2 \) and 1 \( \text{H}_2\text{O} \).
- Balance H: \( \text{Fe(OH)}_2 \) has 2 H, which comes from 1 \( \text{H}_2\text{O} \).
Balanced Equation: \( \text{Fe} + \text{O}_2 + \text{H}_2\text{O} \rightarrow \text{Fe(OH)}_2 \)
---
10. \( \text{Al} + \text{HCl} \rightarrow \text{AlCl}_3 + \text{H}_2 \)
- Start with 1 Al.
- Each \( \text{AlCl}_3 \) has 1 Al, so we need 1 Al.
- Balance Cl: \( \text{AlCl}_3 \) has 3 Cl, so we need 3 HCl.
- Balance H: 3 HCl gives 3 H, which forms 1.5 \( \text{H}_2 \). Multiply everything by 2 to eliminate the fraction.
Balanced Equation: \( 2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2 \)
---
11. \( \text{NH}_3 + \text{O}_2 \rightarrow \text{N}_2\text{O}_3 + \text{H}_2\text{O} \)
- Start with 1 N in \( \text{NH}_3 \).
- Each \( \text{N}_2\text{O}_3 \) has 2 N, so we need 2 \( \text{NH}_3 \).
- Balance H: 2 \( \text{NH}_3 \) have 6 H, so we need 3 \( \text{H}_2\text{O} \).
- Balance O: \( \text{N}_2\text{O}_3 \) has 3 O, and 3 \( \text{H}_2\text{O} \) have 3 O, so we need 3 O from \( \text{O}_2 \).
Balanced Equation: \( 2\text{NH}_3 + 3\text{O}_2 \rightarrow \text{N}_2\text{O}_3 + 3\text{H}_2\text{O} \)
---
12. \( \text{K}_2\text{O}_2 + \text{H}_2\text{O} \rightarrow \text{KOH} + \text{O}_2 \)
- Start with 2 K in \( \text{K}_2\text{O}_2 \).
- Each \( \text{KOH} \) has 1 K, so we need 2 \( \text{KOH} \).
- Balance O: \( \text{K}_2\text{O}_2 \) has 2 O, and 2 \( \text{KOH} \) have 2 O, so we need 1 \( \text{O}_2 \).
- Balance H: 2 \( \text{KOH} \) require 2 H, so we need 1 \( \text{H}_2\text{O} \).
Balanced Equation: \( \text{K}_2\text{O}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{KOH} + \text{H}_2 + \text{O}_2 \)
---
13. \( \text{Mn} + \text{CuCl} \rightarrow \text{Cu} + \text{MnCl}_2 \)
- Start with 1 Mn.
- Each \( \text{MnCl}_2 \) has 1 Mn, so we need 1 Mn.
- Balance Cu: 1 Cu in \( \text{CuCl} \) matches 1 Cu on the right.
- Balance Cl: \( \text{CuCl} \) has 1 Cl, so we need 2 \( \text{CuCl} \) to get 2 Cl for \( \text{MnCl}_2 \).
Balanced Equation: \( \text{Mn} + 2\text{CuCl} \rightarrow 2\text{Cu} + \text{MnCl}_2 \)
---
14. \( \text{Mg(OH)}_2 + \text{H}_3\text{PO}_4 \rightarrow \text{H}_2\text{O} + \text{Mg}_3(\text{PO}_4)_2 \)
- Start with 1 Mg in \( \text{Mg(OH)}_2 \).
- Each \( \text{Mg}_3(\text{PO}_4)_2 \) has 3 Mg, so we need 3 \( \text{Mg(OH)}_2 \).
- Balance P: \( \text{H}_3\text{PO}_4 \) has 1 P, so we need 2 \( \text{H}_3\text{PO}_4 \) to get 2 P for \( \text{Mg}_3(\text{PO}_4)_2 \).
- Balance O: 3 \( \text{Mg(OH)}_2 \) have 6 O, and 2 \( \text{H}_3\text{PO}_4 \) have 8 O, so we need 6 \( \text{H}_2\text{O} \) to balance the O.
- Balance H: 3 \( \text{Mg(OH)}_2 \) have 6 H, and 2 \( \text{H}_3\text{PO}_4 \) have 6 H, which matches 6 H in 6 \( \text{H}_2\text{O} \).
Balanced Equation: \( 3\text{Mg(OH)}_2 + 2\text{H}_3\text{PO}_4 \rightarrow 6\text{H}_2\text{O} + \text{Mg}_3(\text{PO}_4)_2 \)
---
15. \( \text{Ba(HCO}_3)_2 \rightarrow \text{BaCO}_3 + \text{H}_2\text{O} + \text{CO}_2 \)
- Start with 1 Ba in \( \text{Ba(HCO}_3)_2 \).
- Each \( \text{BaCO}_3 \) has 1 Ba, so we need 1 Ba.
- Balance C: \( \text{Ba(HCO}_3)_2 \) has 2 C, so we need 2 \( \text{CO}_2 \).
- Balance O: \( \text{Ba(HCO}_3)_2 \) has 6 O in \( \text{CO}_3 \) groups, 2 O in \( \text{BaCO}_3 \), and 2 O in 2 \( \text{CO}_2 \), so we need 1 \( \text{H}_2\text{O} \).
- Balance H: \( \text{Ba(HCO}_3)_2 \) has 2 H, which matches 2 H in \( \text{H}_2\text{O} \).
Balanced Equation: \( \text{Ba(HCO}_3)_2 \rightarrow \text{BaCO}_3 + \text{H}_2\text{O} + 2\text{CO}_2 \)
---
16. \( \text{Zn} + \text{S}_8 \rightarrow \text{ZnS} \)
- Start with 1 Zn.
- Each \( \text{ZnS} \) has 1 Zn, so we need 1 Zn.
- Balance S: \( \text{S}_8 \) has 8 S, so we need 8 \( \text{ZnS} \).
- Balance Zn: 8 \( \text{ZnS} \) require 8 Zn.
Balanced Equation: \( 8\text{Zn} + \text{S}_8 \rightarrow 8\text{ZnS} \)
---
17. \( \text{LiHCO}_3 \rightarrow \text{Li}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \)
- Start with 1 Li in \( \text{LiHCO}_3 \).
- Each \( \text{Li}_2\text{CO}_3 \) has 2 Li, so we need 2 \( \text{LiHCO}_3 \).
- Balance C: 2 \( \text{LiHCO}_3 \) have 2 C, so we need 1 \( \text{CO}_2 \) and 1 \( \text{Li}_2\text{CO}_3 \).
- Balance O: 2 \( \text{LiHCO}_3 \) have 6 O, and \( \text{Li}_2\text{CO}_3 \) has 3 O, \( \text{CO}_2 \) has 2 O, so we need 1 \( \text{H}_2\text{O} \).
- Balance H: 2 \( \text{LiHCO}_3 \) have 2 H, which matches 2 H in \( \text{H}_2\text{O} \).
Balanced Equation: \( 2\text{LiHCO}_3 \rightarrow \text{Li}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \)
---
18. \( \text{N}_2 + \text{O}_2 \rightarrow \text{N}_2\text{O}_5 \)
- Start with 2 N in \( \text{N}_2 \).
- Each \( \text{N}_2\text{O}_5 \) has 2 N, so we need 1 \( \text{N}_2 \).
- Balance O: \( \text{N}_2\text{O}_5 \) has 5 O, so we need \( \frac{5}{2} \) \( \text{O}_2 \). Multiply everything by 2 to eliminate the fraction.
Balanced Equation: \( 2\text{N}_2 + 5\text{O}_2 \rightarrow 2\text{N}_2\text{O}_5 \)
---
19. \( \text{C}_5\text{H}_{12} + \text{O}_2 \rightarrow \text{H}_2\text{O} + \text{CO}_2 \)
- Start with 5 C in \( \text{C}_5\text{H}_{12} \).
- Each \( \text{CO}_2 \) has 1 C, so we need 5 \( \text{CO}_2 \).
- Balance H: \( \text{C}_5\text{H}_{12} \) has 12 H, so we need 6 \( \text{H}_2\text{O} \).
- Balance O: 5 \( \text{CO}_2 \) have 10 O, and 6 \( \text{H}_2\text{O} \) have 6 O, so we need 8 \( \text{O}_2 \).
Balanced Equation: \( \text{C}_5\text{H}_{12} + 8\text{O}_2 \rightarrow 6\text{H}_2\text{O} + 5\text{CO}_2 \)
---
20. \( \text{C}_4\text{H}_8\text{O}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O} + \text{CO}_2 \)
- Start with 4 C in \( \text{C}_4\text{H}_8\text{O}_2 \).
- Each \( \text{CO}_2 \) has 1 C, so we need 4 \( \text{CO}_2 \).
- Balance H: \( \text{C}_4\text{H}_8\text{O}_2 \) has 8 H, so we need 4 \( \text{H}_2\text{O} \).
- Balance O: \( \text{C}_4\text{H}_8\text{O}_2 \) has 2 O, 4 \( \text{CO}_2 \) have 8 O, and 4 \( \text{H}_2\text{O} \) have 4 O, so we need 5 \( \text{O}_2 \).
Balanced Equation: \( \text{C}_4\text{H}_8\text{O}_2 + 5\text{O}_2 \rightarrow 4\text{H}_2\text{O} + 4\text{CO}_2 \)
---
Final Answer:
\[
\boxed{
\begin{aligned}
1. & \quad \text{Not possible as written.} \\
2. & \quad \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \\
3. & \quad \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \\
4. & \quad 2\text{Al} + \text{Cr}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + 2\text{Cr} \\
5. & \quad 2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2 \\
6. & \quad 2\text{BF}_3 + 3\text{H}_2\text{O} \rightarrow \text{B}_2\text{O}_3 + 6\text{HF} \\
7. & \quad \text{PCl}_5 + \text{AsF}_3 \rightarrow \text{PF}_5 + \text{AsCl}_3 \\
8. & \quad 2\text{K} + \text{O}_2 \rightarrow \text{K}_2\text{O}_2 \\
9. & \quad \text{Fe} + \text{O}_2 + \text{H}_2\text{O} \rightarrow \text{Fe(OH)}_2 \\
10. & \quad 2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2 \\
11. & \quad 2\text{NH}_3 + 3\text{O}_2 \rightarrow \text{N}_2\text{O}_3 + 3\text{H}_2\text{O} \\
12. & \quad \text{K}_2\text{O}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{KOH} + \text{H}_2 + \text{O}_2 \\
13. & \quad \text{Mn} + 2\text{CuCl} \rightarrow 2\text{Cu} + \text{MnCl}_2 \\
14. & \quad 3\text{Mg(OH)}_2 + 2\text{H}_3\text{PO}_4 \rightarrow 6\text{H}_2\text{O} + \text{Mg}_3(\text{PO}_4)_2 \\
15. & \quad \text{Ba(HCO}_3)_2 \rightarrow \text{BaCO}_3 + \text{H}_2\text{O} + 2\text{CO}_2 \\
16. & \quad 8\text{Zn} + \text{S}_8 \rightarrow 8\text{ZnS} \\
17. & \quad 2\text{LiHCO}_3 \rightarrow \text{Li}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \\
18. & \quad 2\text{N}_2 + 5\text{O}_2 \rightarrow 2\text{N}_2\text{O}_5 \\
19. & \quad \text{C}_5\text{H}_{12} + 8\text{O}_2 \rightarrow 6\text{H}_2\text{O} + 5\text{CO}_2 \\
20. & \quad \text{C}_4\text{H}_8\text{O}_2 + 5\text{O}_2 \rightarrow 4\text{H}_2\text{O} + 4\text{CO}_2 \\
\end{aligned}
}
\]
Parent Tip: Review the logic above to help your child master the concept of law of conservation of matter worksheet.
