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 c m_{co₃} = \boxed{?} g submit request answer part d 101 g na₂co₃ m_{nacl} = \boxed{?} g

Explanation:

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

Molar mass of \( \text{Na}_2\text{CO}_3 \) is \( 2\times22.99 + 12.01 + 3\times16.00 = 105.99 \, \text{g/mol} \). Moles \( n = \frac{\text{mass}}{\text{molar mass}} = \frac{101 \, \text{g}}{105.99 \, \text{g/mol}} \approx 0.9529 \, \text{mol} \).

Step2: Use stoichiometry for \( \text{NaCl} \)

From reaction: \( 1 \, \text{mol} \, \text{Na}_2\text{CO}_3
ightarrow 2 \, \text{mol} \, \text{NaCl} \). So moles of \( \text{NaCl} = 2 \times 0.9529 \, \text{mol} = 1.9058 \, \text{mol} \).

Step3: Calculate mass of \( \text{NaCl} \)

Molar mass of \( \text{NaCl} \) is \( 22.99 + 35.45 = 58.44 \, \text{g/mol} \). Mass \( m = n \times \text{molar mass} = 1.9058 \, \text{mol} \times 58.44 \, \text{g/mol} \approx 111 \, \text{g} \).

Answer:

\( \approx 111 \, \text{g} \)