Question

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

Answer 1

The mass of water is calculated as follows

find the moles of each reagent

that is moles = mass/molar mass

for H2s = 84.7/ 34= 2.485 moles
O2 = 78.4 / 32 = 2.45 moles

since 2 moles of H2S react with 3 moles of O2 therefore 2.45 moles of oxygen will be used up therefore O2 is the limiting reagent and H2S is in excess

2H2S + 3O2 ----->2So2 + 2H2O

by use of mole ratio between O2 and H20 which is 3:2 the moles of H2O is therefore = 2.45 x2/3= 1.63 moles of H2O

mass of H2O = moles x molar mass

= 1.63 g x 18g/mol = 29.4 g