Problem: 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.        

FREE Expert Solution

Reaction: CaCO3(aq) + HCl(aq) → CaCl2(aq) + H2CO3(aq)

H2CO3 decomposes to H2O(l) + CO2(g)


Reaction: CaCO3(aq) + 2 HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

Calculate the moles HCl:

moles HCl=30.00 mL×10-3 L1 mL×1.035 molL

moles HCl = 0.03105 mol


Calculate the moles excess HCl:

HCl + NaOH → NaCl + H2O

moles excess acid = moles NaOH used for neutralization

moles excess HCl=11.56 mL×10-3 L1 mL×1.010 molL

moles excess HCl = 0.0116756 mol

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Problem Details

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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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Equivalence Point concept. If you need more Equivalence Point practice, you can also practice Equivalence Point practice problems.