The 1-Question Final Exam

A few people have asked for the answer to this question, so......

325.0mL of 0.825M sodium carbonate solution reacts with 11.348L of hydrogen chloride gas at 1.082atm pressure and 17.28°C to produce carbon dioxide gas, liquid water, and sodium chloride. How many grams of carbon dioxide gas can be produced by this reaction? If all of the sodium chloride that can be produced by this reaction is dissolved in enough water to make 450.0mL of solution, what will be the concentration of sodium chloride in the solution?

First, we need a balanced chemical equation. This is a gas forming reaction.

Na₂CO₃(aq) + 2 HCl(g) --> H₂O(l) + CO₂(g) + 2 NaCl(aq)
{Remember, carbonic acid spontaneously decomposes to water and carbon dioxide gas.}

Now we've got a couple stoichiometry problems...

(0.825 mols Na₂CO₃/L sol'n)(0.3250L sol'n)(^{1mol CO₂(g)}/_{1mol Na2CO3(aq)})(44.009 ^{g CO₂}/_{mol CO2} ) = 11.8g CO₂(g)
(^{(1.082atm)(11.348L HCl(g))}/_{(0.08206 ^L.atm/mol.K)(290.43K)})(^{1mol CO₂(g)}/_{2mol HCl(g)})(44.009 ^{g CO₂}/_{mol CO2} ) = 11.3g CO₂(g)
Since the HCl(g) produces less product, it must be the limiting reagent, so this reaction can produce 11.3g of CO₂(g).

Knowing that HCl(g) is the limiting reagent, we can calculate the amount of NaCl produced by this reaction:
(^{(1.082atm)(11.348L HCl(g))}/_{(0.08206 ^L.atm/mol.K)(290.43K)})(^{2mol NaCl(aq)}/_{2mol HCl(g)}) = 0.515197 mols NaCl(aq) {NOTE: yes, this is too many sig figs, but it's the middle of a problem so I do not want to round off to the correct number of sig figs yet...}

If this amount of NaCl is dissolved in enough water to make 450.0mL of solution, the concentration is:

0.515197mols NaCl / 0.4500L sol'n = 1.145M NaCl(aq)