Question

missed this? watch kcv: reaction stoichiometry, iwe: mass-to-mass conversions. read section 8.4. you can click on the review link to access the section in your etext. for the reaction shown, calculate how many grams of each product form when the following amounts of reactant completely react to form products. assume that there is more than enough of the other reactant. 2 hcl(aq) + na₂co₃(aq) → 2 nacl(aq) + h₂o(l) + co₂(g) part d 10.1 g na₂co₃ m_{nacl} = g submit request answer

Explanation:

Step1: Find molar mass of \( \text{Na}_2\text{CO}_3 \)

Molar mass of \( \text{Na} = 22.99 \, \text{g/mol} \), \( \text{C} = 12.01 \, \text{g/mol} \), \( \text{O} = 16.00 \, \text{g/mol} \).
Molar mass of \( \text{Na}_2\text{CO}_3 = 2(22.99) + 12.01 + 3(16.00) = 105.99 \, \text{g/mol} \).

Step2: Calculate moles of \( \text{Na}_2\text{CO}_3 \)

Moles \( = \frac{\text{mass}}{\text{molar mass}} = \frac{10.1 \, \text{g}}{105.99 \, \text{g/mol}} \approx 0.09529 \, \text{mol} \).

Step3: Use stoichiometry to find moles of \( \text{NaCl} \)

From the reaction: \( 2 \, \text{HCl} + \text{Na}_2\text{CO}_3
ightarrow 2 \, \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \), the mole ratio of \( \text{Na}_2\text{CO}_3 \) to \( \text{NaCl} \) is \( 1:2 \).
Moles of \( \text{NaCl} = 2 \times 0.09529 \, \text{mol} = 0.19058 \, \text{mol} \).

Step4: Calculate molar mass of \( \text{NaCl} \)

Molar mass of \( \text{NaCl} = 22.99 + 35.45 = 58.44 \, \text{g/mol} \).

Step5: Find mass of \( \text{NaCl} \)

Mass \( = \text{moles} \times \text{molar mass} = 0.19058 \, \text{mol} \times 58.44 \, \text{g/mol} \approx 11.1 \, \text{g} \).

Answer:

\( 11.1 \, \text{g} \) (rounded to three significant figures)