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Question

Use the following balanced reaction:

Na₂CO₃ + Ca(HC₂H₃O₂)₂ —> 2NaHC₂H₃O₂ + CaCO₃

If you have 7.95 moles of Na₂CO₃ and 9.20 moles of Ca(HC₂H₃O₂)₂, how many moles of NaHC₂H₃O₂ will be produced?

_____ mole NaHC₂H₃O₂

Answer 1

Na₂CO₃ + Ca(HC₂H₃O₂)₂ —> 2NaHC₂H₃O₂ + CaCO₃ ... balanced equation

Whenever you are given the amounts of BOTH reactants, the first thing you should do is find which, if any, is limiting.

Looking at the balance equation, it is clear that Na₂CO₃ and Ca(HC₂H₃O₂)₂ react in a 1:1 mole ratio, so the reactant that is present in the smaller amount will be limiting, and that happens to be Na₂CO₃ (7.95 mol vs. 9.20 mol).

Now, using 7 .95 mol Na₂CO₃, we can determine the moles of NaHC₂H₃O₂ that can be produced:

7.95 mol Na₂CO₃ x 2 moles NaHC₂H₃O₂ / mole Na₂CO₃ = 15.9 moles NaHC₂H₃O₂