Question

for the following reaction, 6.50 grams of sulfur dioxide are mixed with excess water. the reaction yields 7.07 grams of sulfurous acid (h₂so₃). so₂(g) + h₂o(l) → h₂so₃(g) what is the theoretical yield of sulfurous acid (h₂so₃)? theoretical yield = g what is the percent yield for this reaction? percent yield = % 3 item attempts remaining use the references to access important values if needed for this question.

Explanation:

Step1: Calculate molar mass of $SO_2$

The molar mass of $SO_2$ ($M_{SO_2}$) is $M_{S}+2\times M_{O}=32.07\ g/mol + 2\times16.00\ g/mol=64.07\ g/mol$.

Step2: Calculate moles of $SO_2$

The moles of $SO_2$ ($n_{SO_2}$) is $n_{SO_2}=\frac{m_{SO_2}}{M_{SO_2}}=\frac{6.50\ g}{64.07\ g/mol}\approx0.1015\ mol$.

Step3: Determine mole - ratio from the balanced equation

The balanced equation is $SO_2(g)+H_2O(l)
ightarrow H_2SO_3(g)$. The mole - ratio of $SO_2$ to $H_2SO_3$ is 1:1. So, the moles of $H_2SO_3$ produced in theory ($n_{H_2SO_3,theo}$) is equal to the moles of $SO_2$, $n_{H_2SO_3,theo}=0.1015\ mol$.

Step4: Calculate molar mass of $H_2SO_3$

The molar mass of $H_2SO_3$ ($M_{H_2SO_3}$) is $2\times M_{H}+M_{S}+3\times M_{O}=2\times1.01\ g/mol+32.07\ g/mol + 3\times16.00\ g/mol = 82.09\ g/mol$.

Step5: Calculate theoretical yield of $H_2SO_3$

The theoretical yield of $H_2SO_3$ ($m_{H_2SO_3,theo}$) is $m_{H_2SO_3,theo}=n_{H_2SO_3,theo}\times M_{H_2SO_3}=0.1015\ mol\times82.09\ g/mol\approx8.33\ g$.

Step6: Calculate percent yield

The percent yield formula is $\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%$. The actual yield is $7.07\ g$ and the theoretical yield is $8.33\ g$. So, $\text{Percent Yield}=\frac{7.07\ g}{8.33\ g}\times100\%\approx84.9\%$.

Answer:

Theoretical yield = $8.33$ g
Percent yield = $84.9\%$