Question

how many milliliters of (0.880\text{ m hcl}) are needed to react with (59.8\text{ g}) of (\text{caco}_3)? be sure your answer has the correct number of significant figures.

(2\text{hcl}(aq) + \text{caco}_3(s)
ightarrow \text{cacl}_2(aq) + \text{co}_2(g) + \text{h}_2\text{o}(l))

Explanation:

Calculate moles of CaCO3

$$ M_{\text{CaCO}_3} = 40.08 + 12.01 + 3 \times 16.00 = 100.09\text{ g/mol} $$

$$ n_{\text{CaCO}_3} = \frac{59.8\text{ g}}{100.09\text{ g/mol}} \approx 0.59746\text{ mol} $$

Calculate moles of HCl required

$$ n_{\text{HCl}} = 2 \times n_{\text{CaCO}_3} = 2 \times 0.59746\text{ mol} = 1.19492\text{ mol} $$

Calculate volume of HCl in milliliters

$$ V_{\text{HCl}} = \frac{n_{\text{HCl}}}{M} = \frac{1.19492\text{ mol}}{0.880\text{ M}} \approx 1.35786\text{ L} $$

$$ V_{\text{HCl}} = 1.35786\text{ L} \times 1000\text{ mL/L} \approx 1357.86\text{ mL} \approx 1.36 \times 10^3\text{ mL} $$

Answer:

How many milliliters of \(0.880\text{ M HCl}\) are needed to react with \(59.8\text{ g}\) of \(\text{CaCO}_3\)? Be sure your answer has the correct number of significant figures.

<blank>\(1.36 \times 10^3\)</blank> mL