Question

consider the following unbalanced equation:H2S(g) + O2(g) —> SO2(g) + H2O(g)determine the maximum number of moles of SO2 produced from 11.0 moles of H2S and 2.00 moles of O2

Answer 1

First we need to balance the redox-equation:

$2H_2S+3O_2\to2SO_2+2H_2O,$

We calculate the number of grams of H2S and O2 using the atomic mass:

$11molesH_2S\cdot\frac{34gH_2S}{1molH_2S}=374gH_2S,$$2molesO_2\cdot\frac{32gO_2}{1molO_2}=64gO_{2.}$

68 g of H2S (2*atomic mass of H2S) produces 96 g of O2 (3*atomic mass of O2):

Then, we calculate the limit reagent and excess reagent, following the calculation:

$374gH_2S\cdot\frac{96gO_2}{68gH_2S}=528gO_{2,}$$64gO_2\cdot\frac{68gH_{2_{}}S}{96gO_2}=45.33gH_2S.$