A student mixes 20.00 mL of 0.190 M Ca(NO3)2 and 12.00 mL of 0.150 M Na2CO3. What is the limiting reactant? Show your work.

Answer 1

Limiting reactant problem.

There are several ways to do limiting reactant problems, but the easiest way to identify the limiting reactant, in my opinion, is to find the moles of each reactant, divide each by the respective coefficient in the balanced equation, and whichever is less is the limiting reactant. For this problem, we proceed as follows:

Write a correctly balanced equation:

Ca(NO₃)₂(aq) + Na₂CO₃(aq) ==> CaCO₃(s) + 2NaNO₃(aq)

Find moles of each and divide by coefficient:

moles Ca(NO₃)₂ = 20.00 ml x 1 L/1000 ml x 0.190 mol/L = 0.0038 mol (÷ 1 = 0.0038)

moles Na₂CO₃ = 12.00 ml x 1 L/1000 ml x 0.150 mol/L = 0.0018 mol (÷ 1 = 0.0018)

Since 0.0018 is less than 0.0038 Na₂CO₃ is limiting.

(NOTE: in this case it was easy to see what is limiting because mole ratio is 1:1)

The longer way to do the same thing is to determine the moles of product (any product) formed from each reactant, and see which one produces the smaller amount. This would be as follows:

For Ca(NO₃)₂:

moles CaCO₃ produced = 20.00 ml Ca(NO₃)₂ x 1L/1000 ml x 0.190 mol/L x 1 mol CaCO₃/mol Ca(NO₃)₂ = 0.0038 moles CaCO₃ produced

For Na₂CO₃:

moles CaCO₃ produced = 12.00 ml x 1L/1000 ml x 0.15 mol/L x 1 mol CaCO₃/mol Na₂CO₃ = 0.0018 moles CaCO₃ produced. Thus, Na₂CO₃ is limiting.