Question

For the following reaction, 3.35 grams of sulfur dioxide are mixed with excess oxygen gas . The reaction yields 2.83 grams of sulfur trioxide . sulfur dioxide ( g ) + oxygen ( g ) sulfur trioxide ( g )

What is the theoretical yield of sulfur trioxide ? grams

What is the percent yield for this reaction ? %

Answer 1

1)

Molar mass of SO2,

MM = 1*MM(S) + 2*MM(O)

= 1*32.07 + 2*16.0

= 64.07 g/mol

mass(SO2)= 3.35 g

number of mol of SO2,

n = mass of SO2/molar mass of SO2

=(3.35 g)/(64.07 g/mol)

= 5.229*10^-2 mol

Balanced chemical equation is:

2 SO2 + O2 ---> 2 SO3 +

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)*0.0523

= 0.0523 mol

mass of SO3 = number of mol * molar mass

= 5.229*10^-2*80.07

= 4.187 g

Answer: 4.19 g

2)

% yield = actual mass*100/theoretical mass

= 2.83*100/4.187

= 67.6 %

Answer: 67.6 %