Question

Consider the reaction: 2SO2(g) + O2(g) → 2SO3(g)

Given an initial mass of 11.72 g SO2, an excess of O2, and assuming that all of the reactant is converted to product(s), and none is lost, calculate the mass (g) of SO3 produced by the reaction.

Answer 1

Molar mass of SO2,
MM = 1*MM(S) + 2*MM(O)
= 1*32.07 + 2*16.0
= 64.07 g/mol


mass(SO2)= 11.72 g

number of mol of SO2,
n = mass of SO2/molar mass of SO2
=(11.72 g)/(64.07 g/mol)
= 0.1829 mol
Balanced chemical equation is:
2 SO2 + O2 ---> 2 SO3 + 2 SCl2


Molar mass of SO3,
MM = 1*MM(S) + 3*MM(O)
= 1*32.07 + 3*16.0
= 80.07 g/mol

According to balanced equation
mol of SO3 formed = (2/2)* moles of SO2
= (2/2)*0.182925
= 0.182925 mol


mass of SO3 = number of mol * molar mass
= 0.1829*80.07
= 14.7 g
Answer: 14.7 g