Question

Sulfur dioxide gas (SO2) and oxygen gas (O2) react to form the liquid product of sulfur trioxide (SO3). How much sulfur dioxide would you need to completely react with 6.00 g O2 such that all reactants could be consumed?

Answer 1

1) Balanced chemical equation:

2SO2 (g) +  O2 (g) -> 2SO3 (l)

2) Molar ratios

2 mol SO2 : 1 mol O2 : 2 mol SO3

3) Convert 6.00 g O2 to moles

number of moles = mass in grams / molar mass

number of moles = 6.00 g / 32 g/mol = 0.1875 mol O2.

4) Use proportions with the molar ratios

=> 2 moles SO2 / 1 mol O2 = x / 0.1875 mol O2

=> x = 0.1875 mol O2 * 2 mol SO2 / 1 mol O2 = 0.375 mol SO2.

5) Convert 0.375 mol SO2 to grams

mass in grams = number of moles * molar mass

molar mass SO2 = 32 g/mol + 2*16 g/mol = 64 g/mol

=> mass SO2 = 0.375 mol * 64 g / mol = 24.0 g

Answer: 24.0 g of SO2 are needed to react completely with 6.00 g O2.