In the contact process, $2 SO X_{2} + O X_{2} 2 SO X_{3}$ over a $V X_{2} O X_{5}$ catalyst with excess $O X_{2}$ . If $25.6 g$ of $SO X_{2}$ is converted with an overall yield of $90%$ , calculate the mass of $SO X_{3}$ formed in grams (atomic masses: S = 32, O = 16). Give the answer to two decimal places.

Numerical answer type, answer in g, tolerance 0.01.

Answer: 28.8 g

Moles of

SO X_{2}

25.6/64 = 0.400

. The ratio

SO X_{2} : SO X_{3}

is 1:1, so at 90% yield the moles of

SO X_{3}

0.400 \times 0.90 = 0.360

. Mass =

0.360 \times 80 = 28.80 g