Will mark as brainliest. 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 explain your answer.

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.

It shows that the experiment is designed properly.