Question

als) 11) 3NO2 + H2O → 2HNO3 + NO28 moles of NO2 are reacted. If there is a percent yield rate of 96% HNO3, how many moles ofHNO3 are produced?

Answer 1

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Step-by-step explanation:

Answer 2

13.09 moles of HNO3 are produced.

Explanation:

1st) It is necessary to balance the chemical equation:

`3NO_2+H_2O\rightarrow2HNO_3+NO`

Now we know that 3 moles of NO2 react with 1 mole of H2O to produce 2 moles of HNO3 and 1 mole of NO.

2nd) The percent yield rate, it is used to demonstrate that the reaction is not 100% efficient. According to the formula of percent yield rate, we have that:

`\%YieldRate=\frac{\text{ Actual yield}}{\text{ Theoretical yield}}*100\%`

So, we know that the percent yield rate for this reaction is 96%.

3rd) With the stoichiometry of the balanced equation, we can calculate the Theoretical yield:

`\begin{gathered} 3\text{ moles NO}_2-2\text{ moles HNO}_3 \\ 28\text{moles NO}_2-x=\frac{28\text{moles NO}_2*2\text{ moles HNO}_3}{3\text{ moles NO}_2} \\ x=18.67\text{ moles HNO}_3 \end{gathered}`

Now we know that with 28 moles of NO2, 18.67 moles of HNO3 are produced if the reaction was 100% efficient.

With the molar mass of NO2 (46g/mol) we can convert moles to grams:

`18.67mol*(46g)/(1mol)=858.82g`

4th) With the 858.82 grams of NO2 and the percent yield rate, we can calculate the Actual yield of this reaction:

`\begin{gathered} \%YieldRate=(ActualYield)/(Theoreticalyield)*100\% \\ 96\%=(ActualYield)/(858.82g)*100\% \\ (96\%*858.82g)/(100\%)=ActualYield \\ 824.47g=ActualYield \end{gathered}`

5th) Now that we know the actual yield of the reaction, we can calculate the moles of HNO3 that are produced, using the molar mass of HNO3 (63.01g/mol):

`824.47g*(1mol)/(63.01g)=13.09mol`

So, 13.09 moles of HNO3 are produced.