Question

Hydrogen sulfide reacts with oxygen to form sulfur dioxide and water according to the following unbalanced reaction: H2S (g) + O2 (g) —> SO2 (g) + H2O(g)
If we start with 36.6 g of Hudrogen Sulfide and 15.4 of oxygen, how many grams of water do we make?

Answer 1

The mole ratio for the reaction=$1:1$

Moles of water produced from$H_2S=$ $32gH_2S\times$ $1mol H_2S\over 34.1gH_2S$ $\times$ $1molH_2O\over 1mol H_2S$ $\times$ $18gH_2O\over 1mol H_2O$ $=16.94gH_2O$

Moles of water produced from $O_2=$

$15.4gH_2O\times$ $1mol O_2\over 16.0gH_2O$ $\times$ $1mol H_2O\over 1mol O_2$ $\times$ $18gH_2O\over 1molH_2O$ $=17.33gH_2O$

Total grams produced in the reaction$=16.94g+17.33g=34.27gH_2O$