Question

In the manufacturing process of sulfuric acid, sulfur dioxide is reacted with oxygen to produce sulfur trioxide. Using the equation, 2SO2 (g) + O2 Imported Asset 2SO3 (g), if 64.06g of sulfur dioxide is given an opportunity to react with an excess of oxygen to produce 75.00 g of sulfur trioxide, what is the percent yield of this reaction? 46.83% 60.25% 75.55% 93.68%

Answer 1

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Explanation:

Answer 2

Answer : The percent yield of the reaction is, 93.68 %

Explanation :  Given,

Mass of $SO_2$ = 64.06 g

Molar mass of $SO_2$ = 64 g/mole

Molar mass of $SO_3$ = 80 g/mole

First we have to calculate the moles of $SO_2$.

$\text{Moles of }SO_2=\frac{\text{Mass of }SO_2}{\text{Molar mass of }SO_2}=\frac{64.06g}{64g/mole}=1.0009mole$

Now we have to calculate the moles of $SO_3$.

The balanced chemical reaction will be,

$2SO_2+O_2\rightarrow 2SO_3$

From the balanced reaction, we conclude that

As, 2 moles of $SO_2$ react to give 2 moles of $SO_3$

So, 1.0009 moles of $SO_2$ react to give 1.0009 moles of $SO_3$

Now we have to calculate the mass of $SO_3$

$\text{Mass of }SO_3=\text{Moles of }SO_3\times \text{Molar mass of }SO_3$

$\text{Mass of }SO_3=(1.0009mole)\times (80g/mole)=80.072g$

The theoretical yield of $SO_3$  = 80.072 g

The actual yield of $SO_3$  = 75.00 g

Now we have to calculate the percent yield of $SO_3$

$\%\text{ yield of the reaction}=\frac{\text{Actual yield of }SO_3}{\text{Theoretical yield of }SO_3}\times 100=\frac{75.00g}{80.072g}\times 100=93.68\%$

Therefore, the percent yield of the reaction is, 93.68 %