Question

Sulfuric acid is produced by the following reaction: 2 SO2 + O2 + 2 H2O --> 2 H2SO4 If 300g of SO2 are mixed with 100 g of O2 and 150 g of H2O and the reaction goes to completion determine: a) the limiting reagent, b) grams of sulfuric acid produced, c) how many grams of each reactant is left over.

Answer 1

change all to mol

mol of SO2 = m/MW = 300/64.066 = 4.68267

mol of O2 = m/MW = 100/32 = 3.125

mol of H2O = m/MW = 150/18 = 8.33333

ratios are

2:1:2

then we have limiting reactant must be SO2 since it need

SO2 = O2 = 3.125*2 = 6.25 mol of SO2 which we do not have

SO2:H2O = 1 so 8.33333 mol of H2O per 8.33333mol ofSO2, we do not have

SO2 is the limitng reactant

b)

since ratio is 2:2 os SO2: H2SO4

then

4.68267 mol of H2SO4 is produced

mass =mol*MW = 4.68267*98 = 458.90166 g of H2SO4

c)

reactant left:

3.125 - 4.68267/2 = 0.783665

mass of O2 = 0.783665*32 = 25.07728 g

8.33333 - 4.68267 = 3.65066

mass of H2O = 3.65066*18 = 65.71188

total mass = 65.71188+25.07728 = 90.78916 g of reactant