posted by maria on .
If 10L of CO2 at STP and 50g sodium chloride react with excess ammonia and water, what is the limiting reactant and the mass of baking soda produced?
CO2 + H2O + NaCl + NH3 ==> NaHCO3 + NH4Cl
mols CO2 = 10 L/22.4 = 0.446
mols NaCl= 50/58.44 = 0.856
Now convert each to mols NaHCO3 separately.
First CO2 AS IF WE had all of the NaCl we needed.
0.446 mol CO2 x (1 mol NaHCO3/1 mol CO2) = 0.446 mol NaHCO3 produced.
Do the same for NaCl AS IF we had all of the CO2 we needed.
0.856 mol NaCl x (1 mol NaHCO3/1 mol NaCl) = 0.856 mol NaHCO3 produced.
You have two different answers; obviously one of them must be wrong. The correct answer in limiting reagent problem is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Therefore, CO2 is the limiting reagent.
Here is a worked example of a limiting reagent problem. Remember the steps. This worked example shows two ways to solve the problem.