Question

Ammonia, carbon dioxide, and water vapor react to form ammonium bicarbonate as follows: NH_3(aq) + CO_2(aq) + H_2O(L) rightarrow NH_4HCO_3(aq) Suppose 50.0 g of NH_3, 80.0 g of CO_2, and 2.00 mol of H_2O are reacted. What is the maximum number of grams of NH_4HCO_3 that can be produced?

Answer 1

Change all to mole, so we can compare

MW of NH3 = 17 g/mol

MW of CO2 = 44 g/mol

Mw of H2O = 18 g/mol

mol of NH3 = mass / MW = 50/17 = 2.94

mol of CO2 = mass / MW = 80/44 = 1.82

mo of H2O = 2

make sure the balance is correct:

it is, (count of C,H,N,O are correct)

so

1:1:1 ratio is correct

1.82 mol of CO2 will limit

therefore

only 1.82 mol of NH4HCO3 can be produced

mass = mol*MW = 1.82*79.056 = 143.88192 g of NH4HCO3 can be produced