Question

Consider the reaction:

2SO_2(g)+O_2(g)--->2SO_3(s)

(a) If 285.5 mL of SO₂ is allowed to react with 158.9mL of O₂ (both measured at STP), what is the limiting reactant and the theoretical yield of SO₃?

(b) If 2.805g of SO₃ is collected (measured at STP), what is the percent yield for the reaction?

Answer 1

Given reaction:

2SO_2(g)+O_2(g)--->2SO_3(s)

Volume of SO2 = 285.5 mL

Volume of O2 =- 158.9 mL

Since reaction condition is STP so use volume directly for the calculation of final volume of product and limiting reactant.

At STP 1 mol of any gas = 22.4 L

From above reaction mole ratio of SO2 and O2 is 2:1

Lets say that 2 L SO2 requires 1 L O2. Lets calculate volume of O2 required to react with 285.5 mL SO3

Volume of O2 = 285.5 x 10^-3 L SO2 x 1 L O2 / 2 L SO2

=0.14275 L

But actual volume of O2 = 0.1589 L so O2 is in excess SO2 is limiting reactant.

Lets calculate volume of product formed from limiting reactant.

Volume ratio of SO3 to SO2 is 2 : 2 or 1 : 1

Volume of SO3 formed = 0.2855 L SO2 x 2 L SO3 / 2 L SO2

= 0.2855 L SO3

Now calculation of mass of SO3

Mass of SO3

= 0.2855 L SO3 x (1 mol SO3/ 22.4 L SO3)x molar mass of SO3

=0.01275 mol SO3 x 80.066 g/mol

= 1.013 g SO3

b). Percent yield = actual mass / theoretical mass ) x 100

=275 %

Mass of actual is greater that theoretical yield.