Grade 10 Science homework sheet focusing on balancing chemical equations, with space for student name, group, and date.

Here are the balanced chemical equations for your homework sheet. I have worked through each one to ensure the number of atoms for every element is the same on both sides.

1. $N_2 + 3H_2 \rightarrow 2NH_3$
*(Nitrogen: 2 on both sides. Hydrogen: $3 \times 2 = 6$ on left, $2 \times 3 = 6$ on right.)*

2. $S_8 + 12O_2 \rightarrow 8SO_3$
*(Sulfur: 8 on both sides. Oxygen: $12 \times 2 = 24$ on left, $8 \times 3 = 24$ on right.)*

3. $2HgO \rightarrow 2Hg + O_2$
*(Mercury: 2 on both sides. Oxygen: 2 on both sides.)*

4. $Zn + 2HCl \rightarrow ZnCl_2 + H_2$
*(Zinc: 1 on both sides. Hydrogen: 2 on both sides. Chlorine: 2 on both sides.)*

5. $SiCl_4 + 4H_2O \rightarrow H_4SiO_4 + 4HCl$
*(Silicon: 1 on both sides. Chlorine: 4 on both sides. Hydrogen: $4 \times 2 = 8$ on left, $4 + 4 = 8$ on right. Oxygen: 4 on both sides.)*

6. $2Na + 2H_2O \rightarrow 2NaOH + H_2$
*(Sodium: 2 on both sides. Hydrogen: $2 \times 2 = 4$ on left, $2 + 2 = 4$ on right. Oxygen: 2 on both sides.)*

7. $2H_3PO_4 \rightarrow H_4P_2O_7 + H_2O$
*(Hydrogen: $2 \times 3 = 6$ on left, $4 + 2 = 6$ on right. Phosphorus: 2 on both sides. Oxygen: $2 \times 4 = 8$ on left, $7 + 1 = 8$ on right.)*

8. $2Si_2H_3 + \frac{11}{2}O_2 \rightarrow 4SiO_2 + 3H_2O$
*To remove the fraction, multiply everything by 2:*
$4Si_2H_3 + 11O_2 \rightarrow 8SiO_2 + 6H_2O$
*(Silicon: 8 on both sides. Hydrogen: $4 \times 3 = 12$ on left, $6 \times 2 = 12$ on right. Oxygen: $11 \times 2 = 22$ on left, $(8 \times 2) + 6 = 22$ on right.)*

9. $2Al(OH)_3 + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 6H_2O$
*(Aluminum: 2 on both sides. Sulfate groups ($SO_4$): 3 on both sides. Hydrogen: $(2 \times 3) + (3 \times 2) = 12$ on left, $6 \times 2 = 12$ on right. Oxygen (in water/hydroxide balance): 6 from hydroxide + 12 from acid = 18 total O? Let's check atoms. Left O: $2(3) + 3(4) = 18$. Right O: $3(4) + 6(1) = 18$. Correct.)*

10. $4Fe + 3O_2 \rightarrow 2Fe_2O_3$
*(Iron: 4 on both sides. Oxygen: $3 \times 2 = 6$ on left, $2 \times 3 = 6$ on right.)*

11. $Fe_2(SO_4)_3 + 6KOH \rightarrow 3K_2SO_4 + 2Fe(OH)_3$
*(Iron: 2 on both sides. Sulfate: 3 on both sides. Potassium: 6 on both sides. Hydroxide: 6 on both sides.)*

12. $4FeS_2 + 11O_2 \rightarrow 2Fe_2O_3 + 8SO_2$
*(Iron: 4 on both sides. Sulfur: $4 \times 2 = 8$ on left, 8 on right. Oxygen: $11 \times 2 = 22$ on left, $(2 \times 3) + (8 \times 2) = 6 + 16 = 22$ on right.)*

13. $2Al + 3FeO \rightarrow Al_2O_3 + 3Fe$
*(Aluminum: 2 on both sides. Iron: 3 on both sides. Oxygen: 3 on both sides.)*

14. $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$
*(Sodium: 2 on both sides. Carbon: 1 on both sides. Oxygen: 3 on left, $1 + 2 = 3$ on right. Hydrogen: 2 on both sides. Chlorine: 2 on both sides.)*

15. $2K + Br_2 \rightarrow 2KBr$
*(Potassium: 2 on both sides. Bromine: 2 on both sides.)*

16. $P_4 + 5O_2 \rightarrow 2P_2O_5$
*(Phosphorus: 4 on both sides. Oxygen: $5 \times 2 = 10$ on left, $2 \times 5 = 10$ on right.)*

17. $2C_2H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2O$
*(Carbon: $2 \times 2 = 4$ on left, 4 on right. Hydrogen: $2 \times 2 = 4$ on left, $2 \times 2 = 4$ on right. Oxygen: $5 \times 2 = 10$ on left, $(4 \times 2) + 2 = 10$ on right.)*

Final Answer:
1. $N_2 + 3H_2 \rightarrow 2NH_3$
2. $S_8 + 12O_2 \rightarrow 8SO_3$
3. $2HgO \rightarrow 2Hg + O_2$
4. $Zn + 2HCl \rightarrow ZnCl_2 + H_2$
5. $SiCl_4 + 4H_2O \rightarrow H_4SiO_4 + 4HCl$
6. $2Na + 2H_2O \rightarrow 2NaOH + H_2$
7. $2H_3PO_4 \rightarrow H_4P_2O_7 + H_2O$
8. $4Si_2H_3 + 11O_2 \rightarrow 8SiO_2 + 6H_2O$
9. $2Al(OH)_3 + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 6H_2O$
10. $4Fe + 3O_2 \rightarrow 2Fe_2O_3$
11. $Fe_2(SO_4)_3 + 6KOH \rightarrow 3K_2SO_4 + 2Fe(OH)_3$
12. $4FeS_2 + 11O_2 \rightarrow 2Fe_2O_3 + 8SO_2$
13. $2Al + 3FeO \rightarrow Al_2O_3 + 3Fe$
14. $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$
15. $2K + Br_2 \rightarrow 2KBr$
16. $P_4 + 5O_2 \rightarrow 2P_2O_5$
17. $2C_2H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2O$