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Calculate the volume of carbon dioxide gas produced, at SLC, when 5.00g of calcium carbonate is added to a solution containing 5.00g of nitric acid.

I understand the basics of how to undertake this question but I can't produce the equation or balance it.

Thanks

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+1: Love to see some happy usage of tags :) – Collin Li♦ Mar 2 at 11:18

2 Answers

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Oh sorry! I read your question wrong before :/

To find the products of this reaction, you look at the old acid + metal reactions.

When a Metal Carbonate reacts with an acid, three products will be produced;

  • Salt
  • Carbon Dioxide
  • Water

In your case: CaCO3 + 2HNO3 → Ca(NO3)2 + CO2 + H2O

(i) The CO3 will become CO2.
(ii) The H will combine with the remaining O and form water (H2O),
(iii) which leaves the Ca2+ and the NO3- to form a salt.

Then it's just a matter of balancing the elements on each side.

You can visit these links to learn more incase you dont understand my answer (:

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thanks alot mate :D – Terrance Mar 2 at 9:06
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Hi ;D My name is Terence too xD

First, you write the reaction:

  • CaCO3 + 2HNO3 → Ca(NO3)2 + CO2 + H2O

You then work out the excess reactant through stoichiometry:

  • n(CaCO3) = .0500 mol will react
  • n(HNO3) = .0397 mol will react

We will therefore use the amount of HNO3 to see how much CO2 we have:

  • n(CO2) = .0397 mol of CO2 is present

We can then use the SLC conditions to find the volume using the equation:

Volume= n x VM

  • n = .0397
  • VM = 24.5

The volume will be .97265 L

The least number of significant figures we used was 2 (Molar mass of hydrogen is 1.0) so the answer will be 0.97L

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