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snow_lady [41]
3 years ago
12

If the percent yield for the following reaction is 75.0%, and 25.0 g of NO₂ are consumed in the reaction, how many grams of nitr

ic acid, HNO₃(aq) are produced?
Chemistry
1 answer:
victus00 [196]3 years ago
7 0

Answer:

17.1195 grams of nitric acid are produced.

Explanation:

3NO_2+H_2O\rightarrow 2HNO_3+NO

Moles of nitrogen dioxide :

\frac{25.0 g}{56 g/mol}=0.5434 mol

According to reaction 3 moles of nitrogen dioxides gives 2 moles of nitric acid.

Then 0.5434 moles of nitrogen dioxides will give:

\frac{2}{3}\times 0.5434 mol=0.3623 mol of nitric acid.

Mass of 0.3623 moles of nitric acid :

0.3623 mol\times 63 g/mol=22.8260 g

Theoretical yield = 22.8260 g

Experimental yield = ?

\%Yield=\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100

75\%=\frac{\text{Experimental yield}}{22.8260 g}

Experimental yield of nitric acid = 17.1195 g

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