Question

Ammonia gas can be prepared by the reaction: CaO(s) + 2NH4Cl → 2NH3(g) + H2O(g) + CaCl2(s). If 112 g of CaO react with 224 g NH4Cl, how many moles of reactants and products are there when the reaction is complete?

Answer 1

CaO moles = mass / molar mass

                  = 112 / 56

                   = 2

NH4Cl moles = mass / molar mass

                       = 224 / 53.5

                       = 4.19

CaO(s) + 2NH4Cl ---------------> 2NH3(g) + H2O(g) + CaCl2(s)

1mol        2 mol

2mol         4.19mol

1 mol CaO -----------------> 2 mol NH4Cl

2 mol CaO -------------------> x mol NH4Cl

x = 2 x 2 /1 = 4 mol NH4Cl needed

but we have 4.19 mol NH4Cl

4 moles consumed and 0.19 moles reamined.

reactants remains = 0.19 mol

CaO(s) + 2NH4Cl ---------------> 2NH3(g) + H2O(g) + CaCl2(s)

1mol                                            2                   1             1

2mol                                           4                   1               1

product moles = 4 + 1 + 1 = 6

total product moles when reaction completed = 6