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For the balanced equation shown below, if 84.7 grams of H2S were reacted with 78.4 grams of O2, how many grams of H2O would be produced?
2H2S + 3O2 → 2SO2 + 2H2O


12.4 grams of water


24.9 grams of water


29.4 grams of water


Please explain your answer.

Respuesta :

Molar mass of H2S = 2M(H)+M(S) = 2*1.0+32.0=34.1 g/mol
Molar mass of O2 = 2*M(O)=2*16.0=32.0 g/mol
Molar mass of H2O = 2M(H) +M(O) = 2*1.0+16.0 = 18.0 g/mol

84.7 g H2S*1 mol H2S/34.1 g H2S ≈2.48 mol H2S
78.4 g O2*1 mol O2/32 g O2 =2.45 mol H2S

                                                2H2S +           3O2 → 2SO2 + 2H2O
from reaction                       2 mol              3 mol
from the problem                2.48 mol      2.45 mol

As we can see O2 is limiting reactant, so we are going to find amount of water using O2 data.

                                         
                                                2H2S +           3O2 → 2SO2 +       2H2O
from reaction                                               3 mol                         2 mol
from problem                                              2.45 mol                    x mol

x=(2.45* 2)/3≈1.63 mol H2O

1.63 mol H2O*18 g H2O/1 mol H2O ≈29.34 g H2O

The answer is   29.4 grams of water.

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