CBSE Class 10 Chemistry Worksheet - Chemical Reactions and ... - Free Printable

Problem: Balance the given chemical equations and identify the reactants and products.

#### Definitions:
1. Reactants: Substances that participate in a chemical reaction and are consumed to form new substances.
2. Products: New substances formed as a result of a chemical reaction.
3. Balanced Chemical Equation: An equation where the number of atoms of each element is the same on both sides of the equation.

---

Solutions:

#### a. \( \_\_\_\_ \text{Zn} (s) + \_\_\_\_ \text{AgNO}_3 (aq) \rightarrow \_\_\_\_ \text{Zn(NO}_3)_2 (aq) + \_\_\_\_ \text{Ag} (s) \)

- Step 1: Identify the elements and their initial counts.
- Reactants: Zn, Ag, N, O
- Products: Zn, Ag, N, O

- Step 2: Balance the equation.
- Start with zinc (Zn): There is 1 Zn on the left and 1 Zn on the right, so Zn is balanced.
- Next, balance silver (Ag): There is 1 Ag on the left and 1 Ag on the right, but we need 2 Ag atoms on the right to match the formula of \(\text{Zn(NO}_3)_2\). So, we need 2 \(\text{AgNO}_3\) molecules on the left.
- Now, check nitrogen (N): There are 2 N atoms on the right (from \(\text{Zn(NO}_3)_2\)), so we need 2 N atoms on the left, which is satisfied by 2 \(\text{AgNO}_3\).
- Finally, check oxygen (O): There are 6 O atoms on the right (2 from \(\text{Zn(NO}_3)_2\)), so we need 6 O atoms on the left, which is satisfied by 2 \(\text{AgNO}_3\).

- Balanced Equation:
\[
\text{Zn} (s) + 2 \text{AgNO}_3 (aq) \rightarrow \text{Zn(NO}_3)_2 (aq) + 2 \text{Ag} (s)
\]

- Reactants: \(\text{Zn} (s)\), \(\text{AgNO}_3 (aq)\)
- Products: \(\text{Zn(NO}_3)_2 (aq)\), \(\text{Ag} (s)\)

---

#### b. \( \_\_\_\_ \text{N}_2 (g) + \_\_\_\_ \text{H}_2 (g) \rightarrow \_\_\_\_ \text{NH}_3 (g) \)

- Step 1: Identify the elements and their initial counts.
- Reactants: N, H
- Products: N, H

- Step 2: Balance the equation.
- Start with nitrogen (N): There are 2 N atoms on the left and 1 N atom on the right. To balance N, we need 2 \(\text{NH}_3\) molecules on the right.
- Now, check hydrogen (H): There are 6 H atoms on the right (from 2 \(\text{NH}_3\)), so we need 6 H atoms on the left. This means we need 3 \(\text{H}_2\) molecules on the left.

- Balanced Equation:
\[
\text{N}_2 (g) + 3 \text{H}_2 (g) \rightarrow 2 \text{NH}_3 (g)
\]

- Reactants: \(\text{N}_2 (g)\), \(\text{H}_2 (g)\)
- Products: \(\text{NH}_3 (g)\)

---

#### c. \( \_\_\_\_ \text{Mg(OH)}_2 (aq) + \_\_\_\_ \text{H}_3\text{PO}_4 (aq) \rightarrow \_\_\_\_ \text{H}_2\text{O} (l) + \_\_\_\_ \text{Mg}_3(\text{PO}_4)_2 (aq) \)

- Step 1: Identify the elements and their initial counts.
- Reactants: Mg, O, H, P
- Products: Mg, O, H, P

- Step 2: Balance the equation.
- Start with magnesium (Mg): There is 1 Mg on the left and 3 Mg on the right. To balance Mg, we need 3 \(\text{Mg(OH)}_2\) molecules on the left.
- Next, balance phosphorus (P): There is 1 P on the left and 2 P on the right. To balance P, we need 2 \(\text{H}_3\text{PO}_4\) molecules on the left.
- Now, check oxygen (O): There are 14 O atoms on the right (6 from \(\text{H}_2\text{O}\) and 8 from \(\text{Mg}_3(\text{PO}_4)_2\)). On the left, there are 14 O atoms (6 from 3 \(\text{Mg(OH)}_2\) and 8 from 2 \(\text{H}_3\text{PO}_4\)).
- Finally, check hydrogen (H): There are 12 H atoms on the right (from 6 \(\text{H}_2\text{O}\)) and 12 H atoms on the left (from 3 \(\text{Mg(OH)}_2\) and 2 \(\text{H}_3\text{PO}_4\)).

- Balanced Equation:
\[
3 \text{Mg(OH)}_2 (aq) + 2 \text{H}_3\text{PO}_4 (aq) \rightarrow 6 \text{H}_2\text{O} (l) + \text{Mg}_3(\text{PO}_4)_2 (aq)
\]

- Reactants: \(\text{Mg(OH)}_2 (aq)\), \(\text{H}_3\text{PO}_4 (aq)\)
- Products: \(\text{H}_2\text{O} (l)\), \(\text{Mg}_3(\text{PO}_4)_2 (aq)\)

---

#### d. \( \_\_\_\_ \text{HNO}_3 (aq) + \_\_\_\_ \text{Ni} (s) \rightarrow \_\_\_\_ \text{Ni(NO}_3)_2 (aq) + \_\_\_\_ \text{H}_2 (g) \)

- Step 1: Identify the elements and their initial counts.
- Reactants: H, N, O, Ni
- Products: Ni, N, O, H

- Step 2: Balance the equation.
- Start with nickel (Ni): There is 1 Ni on the left and 1 Ni on the right, so Ni is balanced.
- Next, balance nitrogen (N): There is 1 N on the left and 2 N on the right. To balance N, we need 2 \(\text{HNO}_3\) molecules on the left.
- Now, check oxygen (O): There are 6 O atoms on the right (from \(\text{Ni(NO}_3)_2\)) and 6 O atoms on the left (from 2 \(\text{HNO}_3\)).
- Finally, check hydrogen (H): There are 2 H atoms on the right (from \(\text{H}_2\)) and 2 H atoms on the left (from 2 \(\text{HNO}_3\)).

- Balanced Equation:
\[
2 \text{HNO}_3 (aq) + \text{Ni} (s) \rightarrow \text{Ni(NO}_3)_2 (aq) + \text{H}_2 (g)
\]

- Reactants: \(\text{HNO}_3 (aq)\), \(\text{Ni} (s)\)
- Products: \(\text{Ni(NO}_3)_2 (aq)\), \(\text{H}_2 (g)\)

---

#### e. \( \_\_\_\_ \text{Ba(HCO}_3)_2 (s) \rightarrow \_\_\_\_ \text{BaCO}_3 (s) + \_\_\_\_ \text{H}_2\text{O} (g) + \_\_\_\_ \text{CO}_2 (g) \)

- Step 1: Identify the elements and their initial counts.
- Reactants: Ba, C, H, O
- Products: Ba, C, H, O

- Step 2: Balance the equation.
- Start with barium (Ba): There is 1 Ba on the left and 1 Ba on the right, so Ba is balanced.
- Next, balance carbon (C): There are 2 C atoms on the left and 1 C atom on the right (from \(\text{BaCO}_3\)). To balance C, we need 1 more \(\text{CO}_2\) molecule on the right.
- Now, check oxygen (O): There are 6 O atoms on the left (from \(\text{Ba(HCO}_3)_2\)) and 5 O atoms on the right (3 from \(\text{BaCO}_3\), 1 from \(\text{H}_2\text{O}\), and 2 from \(\text{CO}_2\)). To balance O, we need 1 more \(\text{H}_2\text{O}\) molecule on the right.
- Finally, check hydrogen (H): There are 2 H atoms on the left and 2 H atoms on the right (from \(\text{H}_2\text{O}\)).

- Balanced Equation:
\[
\text{Ba(HCO}_3)_2 (s) \rightarrow \text{BaCO}_3 (s) + \text{H}_2\text{O} (g) + \text{CO}_2 (g)
\]

- Reactants: \(\text{Ba(HCO}_3)_2 (s)\)
- Products: \(\text{BaCO}_3 (s)\), \(\text{H}_2\text{O} (g)\), \(\text{CO}_2 (g)\)

---

#### f. \( \_\_\_\_ \text{BaCl}_2 (aq) + \_\_\_\_ \text{Na}_2\text{SO}_4 (aq) \rightarrow \_\_\_\_ \text{NaCl} (aq) + \_\_\_\_ \text{BaSO}_4 (s) \)

- Step 1: Identify the elements and their initial counts.
- Reactants: Ba, Cl, Na, S, O
- Products: Ba, Cl, Na, S, O

- Step 2: Balance the equation.
- Start with barium (Ba): There is 1 Ba on the left and 1 Ba on the right, so Ba is balanced.
- Next, balance chlorine (Cl): There are 2 Cl atoms on the left and 1 Cl atom on the right. To balance Cl, we need 2 \(\text{NaCl}\) molecules on the right.
- Now, check sodium (Na): There are 2 Na atoms on the right and 2 Na atoms on the left (from \(\text{Na}_2\text{SO}_4\)), so Na is balanced.
- Finally, check sulfur (S) and oxygen (O): There is 1 S atom and 4 O atoms on the left and right, so S and O are balanced.

- Balanced Equation:
\[
\text{BaCl}_2 (aq) + \text{Na}_2\text{SO}_4 (aq) \rightarrow 2 \text{NaCl} (aq) + \text{BaSO}_4 (s)
\]

- Reactants: \(\text{BaCl}_2 (aq)\), \(\text{Na}_2\text{SO}_4 (aq)\)
- Products: \(\text{NaCl} (aq)\), \(\text{BaSO}_4 (s)\)

---

#### g. \( \_\_\_\_ \text{Al}_2(\text{CO}_3)_3 (s) \rightarrow \_\_\_\_ \text{Al}_2\text{O}_3 (s) + \_\_\_\_ \text{CO}_2 (g) \)

- Step 1: Identify the elements and their initial counts.
- Reactants: Al, C, O
- Products: Al, C, O

- Step 2: Balance the equation.
- Start with aluminum (Al): There are 2 Al atoms on the left and 2 Al atoms on the right, so Al is balanced.
- Next, balance carbon (C): There are 3 C atoms on the left and 1 C atom on the right. To balance C, we need 3 \(\text{CO}_2\) molecules on the right.
- Finally, check oxygen (O): There are 9 O atoms on the left and 9 O atoms on the right (3 from \(\text{Al}_2\text{O}_3\) and 6 from 3 \(\text{CO}_2\)).

- Balanced Equation:
\[
\text{Al}_2(\text{CO}_3)_3 (s) \rightarrow \text{Al}_2\text{O}_3 (s) + 3 \text{CO}_2 (g)
\]

- Reactants: \(\text{Al}_2(\text{CO}_3)_3 (s)\)
- Products: \(\text{Al}_2\text{O}_3 (s)\), \(\text{CO}_2 (g)\)

---

#### h. \( \_\_\_\_ \text{Ca} (s) + \_\_\_\_ \text{H}_2\text{O} (l) \rightarrow \_\_\_\_ \text{Ca(OH)}_2 (aq) + \_\_\_\_ \text{H}_2 (g) \)

- Step 1: Identify the elements and their initial counts.
- Reactants: Ca, H, O
- Products: Ca, H, O

- Step 2: Balance the equation.
- Start with calcium (Ca): There is 1 Ca on the left and 1 Ca on the right, so Ca is balanced.
- Next, balance hydrogen (H): There are 2 H atoms on the left and 2 H atoms on the right (from \(\text{H}_2\)), so H is balanced.
- Finally, check oxygen (O): There are 2 O atoms on the left and 2 O atoms on the right (from \(\text{Ca(OH)}_2\)), so O is balanced.

- Balanced Equation:
\[
\text{Ca} (s) + 2 \text{H}_2\text{O} (l) \rightarrow \text{Ca(OH)}_2 (aq) + \text{H}_2 (g)
\]

- Reactants: \(\text{Ca} (s)\), \(\text{H}_2\text{O} (l)\)
- Products: \(\text{Ca(OH)}_2 (aq)\), \(\text{H}_2 (g)\)

---

#### i. \( \_\_\_\_ \text{LiHCO}_3 (s) \rightarrow \_\_\_\_ \text{Li}_2\text{CO}_3 (s) + \_\_\_\_ \text{H}_2\text{O} (g) + \_\_\_\_ \text{CO}_2 (g) \)

- Step 1: Identify the elements and their initial counts.
- Reactants: Li, C, H, O
- Products: Li, C, H, O

- Step 2: Balance the equation.
- Start with lithium (Li): There is 1 Li on the left and 2 Li atoms on the right. To balance Li, we need 2 \(\text{LiHCO}_3\) molecules on the left.
- Next, balance carbon (C): There is 1 C atom on the left and 1 C atom on the right, so C is balanced.
- Now, check hydrogen (H): There are 2 H atoms on the left and 2 H atoms on the right (from \(\text{H}_2\text{O}\)), so H is balanced.
- Finally, check oxygen (O): There are 6 O atoms on the left (from 2 \(\text{LiHCO}_3\)) and 6 O atoms on the right (3 from \(\text{Li}_2\text{CO}_3\), 1 from \(\text{H}_2\text{O}\), and 2 from \(\text{CO}_2\)).

- Balanced Equation:
\[
2 \text{LiHCO}_3 (s) \rightarrow \text{Li}_2\text{CO}_3 (s) + \text{H}_2\text{O} (g) + \text{CO}_2 (g)
\]

- Reactants: \(\text{LiHCO}_3 (s)\)
- Products: \(\text{Li}_2\text{CO}_3 (s)\), \(\text{H}_2\text{O} (g)\), \(\text{CO}_2 (g)\)

---

#### j. \( \_\_\_\_ \text{N}_2 (g) + \_\_\_\_ \text{O}_2 (g) \rightarrow \_\_\_\_ \text{N}_2\text{O}_5 (g) \)

- Step 1: Identify the elements and their initial counts.
- Reactants: N, O
- Products: N, O

- Step 2: Balance the equation.
- Start with nitrogen (N): There are 2 N atoms on the left and 2 N atoms on the right, so N is balanced.
- Next, balance oxygen (O): There are 2 O atoms on the left and 5 O atoms on the right. To balance O, we need 2.5 \(\text{O}_2\) molecules on the left. However, since we cannot have fractional coefficients, we multiply the entire equation by 2 to eliminate the fraction.

- Balanced Equation:
\[
2 \text{N}_2 (g) + 5 \text{O}_2 (g) \rightarrow 2 \text{N}_2\text{O}_5 (g)
\]

- Reactants: \(\text{N}_2 (g)\), \(\text{O}_2 (g)\)
- Products: \(\text{N}_2\text{O}_5 (g)\)

---

#### k. \( \_\_\_\_ \text{MgBr}_2 (aq) + \_\_\_\_ \text{KOH} (aq) \rightarrow \_\_\_\_ \text{KBr} (aq) + \_\_\_\_ \text{Mg(OH)}_2 (s) \)

- Step 1: Identify the elements and their initial counts.
- Reactants: Mg, Br, K, O, H
- Products: Mg, Br, K, O, H

- Step 2: Balance the equation.
- Start with magnesium (Mg): There is 1 Mg on the left and 1 Mg on the right, so Mg is balanced.
- Next, balance bromine (Br): There are 2 Br atoms on the left and 1 Br atom on the right. To balance Br, we need 2 \(\text{KBr}\) molecules on the right.
- Now, check potassium (K): There are 2 K atoms on the right and 1 K atom on the left. To balance K, we need 2 \(\text{KOH}\) molecules on the left.
- Finally, check oxygen (O) and hydrogen (H): There are 2 O atoms and 2 H atoms on the left (from 2 \(\text{KOH}\)) and 2 O atoms and 2 H atoms on the right (from \(\text{Mg(OH)}_2\)).

- Balanced Equation:
\[
\text{MgBr}_2 (aq) + 2 \text{KOH} (aq) \rightarrow 2 \text{KBr} (aq) + \text{Mg(OH)}_2 (s)
\]

- Reactants: \(\text{MgBr}_2 (aq)\), \(\text{KOH} (aq)\)
- Products: \(\text{KBr} (aq)\), \(\text{Mg(OH)}_2 (s)\)

---

#### l. \( \_\_\_\_ \text{Mn} (s) + \_\_\_\_ \text{CuCl}_2 (aq) \rightarrow \_\_\_\_ \text{Cu} (s) + \_\_\_\_ \text{MnCl}_2 (s) \)

- Step 1: Identify the elements and their initial counts.
- Reactants: Mn, Cu, Cl
- Products: Mn, Cu, Cl

- Step 2: Balance the equation.
- Start with manganese (Mn): There is 1 Mn on the left and 1 Mn on the right, so Mn is balanced.
- Next, balance copper (Cu): There is 1 Cu on the left and 1 Cu on the right, so Cu is balanced.
- Finally, balance chlorine (Cl): There are 2 Cl atoms on the left and 2 Cl atoms on the right, so Cl is balanced.

- Balanced Equation:
\[
\text{Mn} (s) + \text{CuCl}_2 (aq) \rightarrow \text{Cu} (s) + \text{MnCl}_2 (s)
\]

- Reactants: \(\text{Mn} (s)\), \(\text{CuCl}_2 (aq)\)
- Products: \(\text{Cu} (s)\), \(\text{MnCl}_2 (s)\)

---

#### m. \( \_\_\_\_ \text{Zn} (s) + \_\_\_\_ \text{S}_8 (s) \rightarrow \_\_\_\_ \text{ZnS} (s) \)

- Step 1: Identify the elements and their initial counts.
- Reactants: Zn, S
- Products: Zn, S

- Step 2: Balance the equation.
- Start with zinc (Zn): There is 1 Zn on the left and 1 Zn on the right, so Zn is balanced.
- Next, balance sulfur (S): There are 8 S atoms on the left and 1 S atom on the right. To balance S, we need 8 \(\text{ZnS}\) molecules on the right.

- Balanced Equation:
\[
8 \text{Zn} (s) + \text{S}_8 (s) \rightarrow 8 \text{ZnS} (s)
\]

- Reactants: \(\text{Zn} (s)\), \(\text{S}_8 (s)\)
- Products: \(\text{ZnS} (s)\)

---

Final Answer:

\[
\boxed{
\begin{aligned}
&\text{a. } \text{Zn} (s) + 2 \text{AgNO}_3 (aq) \rightarrow \text{Zn(NO}_3)_2 (aq) + 2 \text{Ag} (s) \\
&\text{b. } \text{N}_2 (g) + 3 \text{H}_2 (g) \rightarrow 2 \text{NH}_3 (g) \\
&\text{c. } 3 \text{Mg(OH)}_2 (aq) + 2 \text{H}_3\text{PO}_4 (aq) \rightarrow 6 \text{H}_2\text{O} (l) + \text{Mg}_3(\text{PO}_4)_2 (aq) \\
&\text{d. } 2 \text{HNO}_3 (aq) + \text{Ni} (s) \rightarrow \text{Ni(NO}_3)_2 (aq) + \text{H}_2 (g) \\
&\text{e. } \text{Ba(HCO}_3)_2 (s) \rightarrow \text{BaCO}_3 (s) + \text{H}_2\text{O} (g) + \text{CO}_2 (g) \\
&\text{f. } \text{BaCl}_2 (aq) + \text{Na}_2\text{SO}_4 (aq) \rightarrow 2 \text{NaCl} (aq) + \text{BaSO}_4 (s) \\
&\text{g. } \text{Al}_2(\text{CO}_3)_3 (s) \rightarrow \text{Al}_2\text{O}_3 (s) + 3 \text{CO}_2 (g) \\
&\text{h. } \text{Ca} (s) + 2 \text{H}_2\text{O} (l) \rightarrow \text{Ca(OH)}_2 (aq) + \text{H}_2 (g) \\
&\text{i. } 2 \text{LiHCO}_3 (s) \rightarrow \text{Li}_2\text{CO}_3 (s) + \text{H}_2\text{O} (g) + \text{CO}_2 (g) \\
&\text{j. } 2 \text{N}_2 (g) + 5 \text{O}_2 (g) \rightarrow 2 \text{N}_2\text{O}_5 (g) \\
&\text{k. } \text{MgBr}_2 (aq) + 2 \text{KOH} (aq) \rightarrow 2 \text{KBr} (aq) + \text{Mg(OH)}_2 (s) \\
&\text{l. } \text{Mn} (s) + \text{CuCl}_2 (aq) \rightarrow \text{Cu} (s) + \text{MnCl}_2 (s) \\
&\text{m. } 8 \text{Zn} (s) + \text{S}_8 (s) \rightarrow 8 \text{ZnS} (s)
\end{aligned}
}
\]