(ii) Complete the following sentences. 84 g of sodium hydrogencarbonate produces ................ g of sodium carbonate and ............... g of water. 4.2 g of sodium hydrogencarbonate produces ................. g of sodium carbonate.

To complete the sentences involving the decomposition of sodium hydrogencarbonate (NaHCO₃), we'll use the following balanced chemical equation: \[ 2 \, \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \] **Step 1: Calculate Molar Masses** - **NaHCO₃ (Sodium Hydrogencarbonate):** - Na: 23 g/mol - H: 1 g/mol - C: 12 g/mol - O₃: 48 g/mol - **Total:** 84 g/mol - **Na₂CO₃ (Sodium Carbonate):** - Na₂: 46 g/mol - C: 12 g/mol - O₃: 48 g/mol - **Total:** 106 g/mol - **H₂O (Water):** - H₂: 2 g/mol - O: 16 g/mol - **Total:** 18 g/mol **Step 2: Stoichiometric Calculations** 1. **For 84 g of NaHCO₃:** - **Moles of NaHCO₃:** \[ \frac{84 \, \text{g}}{84 \, \text{g/mol}} = 1 \, \text{mol} \] - **From the balanced equation:** - 2 mol NaHCO₃ → 1 mol Na₂CO₃ - 2 mol NaHCO₃ → 1 mol H₂O - **Thus:** - **Moles of Na₂CO₃ produced:** \[ \frac{1 \, \text{mol}}{2} = 0.5 \, \text{mol} \] \[ 0.5 \, \text{mol} \times 106 \, \text{g/mol} = 53 \, \text{g} \] - **Moles of H₂O produced:** \[ \frac{1 \, \text{mol}}{2} = 0.5 \, \text{mol} \] \[ 0.5 \, \text{mol} \times 18 \, \text{g/mol} = 9 \, \text{g} \] 2. **For 4.2 g of NaHCO₃:** - **Moles of NaHCO₃:** \[ \frac{4.2 \, \text{g}}{84 \, \text{g/mol}} = 0.05 \, \text{mol} \] - **Moles of Na₂CO₃ produced:** \[ 0.05 \, \text{mol} \times \frac{1}{2} = 0.025 \, \text{mol} \] \[ 0.025 \, \text{mol} \times 106 \, \text{g/mol} = 2.65 \, \text{g} \] **Final Completed Sentences:** 1. **84 g of sodium hydrogencarbonate produces _53 g_ of sodium carbonate and _9 g_ of water.** 2. **4.2 g of sodium hydrogencarbonate produces _2.65 g_ of sodium carbonate.**