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.
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.
