Question

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!!

Answer 1

Equation is already balanced

SO2+O2+H2O—>H2SO4

Step-by-step explanation:

Answer 2

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`SO_2(g)+(1)/(2) O_2(g)+H_2O(l)\rightarrow H_2SO_4(l)`

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`m_(H_2SO_4)=212.6gH_2SO_4`

*
`Y=94.1\%`

Step-by-step explanation:

Hello,

In this case, the balanced chemical reaction is:

`SO_2(g)+(1)/(2) O_2(g)+H_2O(l)\rightarrow H_2SO_4(l)`

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

`(n_1)/(V_1)= (n_2)/(V_2)\\\\n_2=(n_1V_2)/(V_1) =(1mol*48.6L)/(22.4L)=2.17molSO_2`

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

`m_(H_2SO_4)=2.17molSO_2*(1molH_2SO_4)/(1molSO_2)*(98gH_2SO_4)/(1molH_2SO_4)=212.6gH_2SO_4`

Finally, the percent yield results:

`Y=(200g)/(212.6g)*100\%=94.1\%`

Best regards.