Reaction: CaCO_{3}(aq) + HCl(aq) → CaCl_{2}(aq) + H_{2}CO_{3}(aq)

H_{2}CO_{3} decomposes to H_{2}O(l) + CO_{2}(g)

Reaction: CaCO_{3}(aq) + 2 HCl(aq) → CaCl_{2}(aq) + H_{2}O(l) + CO_{2}(g)

**Calculate the moles HCl:**

$\mathbf{moles}\mathbf{}\mathbf{HCl}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{00}\mathbf{}\overline{)\mathbf{mL}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\overline{)\mathbf{L}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{.}\mathbf{035}\mathbf{}\mathbf{mol}}{\overline{)\mathbf{L}}}$

**moles HCl = 0.03105 mol**

**Calculate the moles excess HCl:**

HCl + NaOH → NaCl + H_{2}O

moles excess acid = moles NaOH used for neutralization

$\mathbf{moles}\mathbf{}\mathbf{excess}\mathbf{}\mathbf{HCl}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{56}\mathbf{}\overline{)\mathbf{mL}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\overline{)\mathbf{L}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{.}\mathbf{010}\mathbf{}\mathbf{mol}}{\overline{)\mathbf{L}}}$

**moles excess HCl = 0.0116756 mol**

A 1.130 g sample of limestone rock is pulverized and then treated with 30.00 mL of 1.035 M HCl solution. The excess acid then requires 11.56 mL of 1.010 M NaOH for neutralization.

You may want to reference (Pages 149 - 153) Section 4.6 while completing this problem.

Calculate the percent by mass of calcium carbonate in the rock, assuming that it is the only substance reacting with the HCl solution.

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