Suppose 50.0 g NH3, 80.0 g CO2, and 2.00 mol H2O are reacted. Determine the

Question

maximum grams of NH4HCO3 that can be produced.

Answer 1

This is a limiting reagent (LR) problem and you know that because amounts are given for all of the reactants instead of just one.
NH3 + CO2 + H2O ==> NH4HCO3

I do these the long way.
1. How much product can we get from 50.0 g NH3 if all of the others are in excess?
mols NH3 = g/molar mass = estimated 3 mol but this is an estimate and you should rdo this and all of the calculations that follow more accurately.
Then 3 mols NH3 x (1 mol NH4HCO3/1 mol NH3) = 3 x 1/1 = 3 mols NH4HCO3.

2. How much product can we get from 80.0 g CO2 if all of the others are in excess?
mols CO2 = g/molar mass = estimated 1.8 mols CO2.
Then 2 mols CO2 x (1 mol NH4HCO3/1 mol CO2) = 1.8 x 1/1 = estimated 1.8 mols NH4HCO3.

3. How much product can we get from 2.00 mls H2O if all of the others are in excess?
2.00 mols H2O x (1 mol NH4HCO3/1 mol H2O) = 2.00 x 1/1 = 2.00 mols NH4HCO3.

You can see that the values for NH4HCO3 vary so only one of those can be right. The correct value in LR problems is ALWAYS the smallest number and the reagent responsible for that number is the LR. So the LR is CO2 and estimated 1.8 mols NH4HCO3 will be formed.

Convert that to grams. grams = mols x molar mass = ?