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Sulfur dioxide gas (SO2) reacts with excess oxygen gas (O2) and excess liquid water (H2O) to form liquid sulfuric acid (H2SO4). In the laboratory, a chemist carries out this reaction with 48.6 L of sulfur dioxide and gets 200 g of sulfuric acid.
• Write a balanced equation for the reaction.
• Calculate the theoretical yield of sulfuric acid.
• Calculate the percent yield of the reaction.
(One mole of any gas occupies 22.4 L under certain conditions of temperature and pressure. Assume those conditions for this question.)

Plz need help ASAP!!

Respuesta :

Answer:

Equation is already balanced

SO2+O2+H2O—>H2SO4

Explanation:

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sebassandin

Answer:

* [tex]SO_2(g)+\frac{1}{2} O_2(g)+H_2O(l)\rightarrow H_2SO_4(l)[/tex]

* [tex]m_{H_2SO_4}=212.6gH_2SO_4[/tex]

* [tex]Y=94.1\%[/tex]

Explanation:

Hello,

In this case, the balanced chemical reaction is:

[tex]SO_2(g)+\frac{1}{2} O_2(g)+H_2O(l)\rightarrow H_2SO_4(l)[/tex]

Now, by means of the Avogadro's law, it is possible to compute the moles of sulfur dioxide as shown below:

[tex]\frac{n_1}{V_1}= \frac{n_2}{V_2}\\\\n_2=\frac{n_1V_2}{V_1} =\frac{1mol*48.6L}{22.4L}=2.17molSO_2[/tex]

Thus, by stoichiometry, the theoretical yield of sulfuric acid in grams is:

[tex]m_{H_2SO_4}=2.17molSO_2*\frac{1molH_2SO_4}{1molSO_2}*\frac{98gH_2SO_4}{1molH_2SO_4}=212.6gH_2SO_4[/tex]

Finally, the percent yield results:

[tex]Y=\frac{200g}{212.6g}*100\%=94.1\%[/tex]

Best regards.

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