question 23 of 31 what is the percent yield for the reaction below when 544.5 g so2 and 160.0 g o2 produce…

Explanation:

Step1: Calculate moles of reactants

Molar mass of $SO_2$ is $64.07\ g/mol$, moles of $SO_2=\frac{544.5\ g}{64.07\ g/mol}\approx8.5\ mol$. Molar mass of $O_2$ is $32\ g/mol$, moles of $O_2 = \frac{160.0\ g}{32\ g/mol}= 5\ mol$.

Step2: Determine limiting reactant

From the balanced equation $2SO_2(g)+O_2(g)\to2SO_3(g)$, the mole - ratio of $SO_2$ to $O_2$ is 2:1. For 5 mol of $O_2$, we need $2\times5 = 10$ mol of $SO_2$, but we have only 8.5 mol of $SO_2$. So, $SO_2$ is the limiting reactant.

Step3: Calculate theoretical yield of $SO_3$

The mole - ratio of $SO_2$ to $SO_3$ is 1:1. So, moles of $SO_3$ produced theoretically from 8.5 mol of $SO_2$ is 8.5 mol. Molar mass of $SO_3$ is $80.07\ g/mol$, theoretical yield of $SO_3=8.5\ mol\times80.07\ g/mol = 680.6\ g$.

Step4: Calculate percent yield

Percent yield = $\frac{\text{actual yield}}{\text{theoretical yield}}\times100%=\frac{382.0\ g}{680.6\ g}\times100%\approx56.1%$

Answer:

C. 56.1%