Answer:
[tex]n_{HNO_3}=0.584molHNO_3[/tex]
Explanation:
Hello.
In this case, given the described reaction, the formation of nitric acid turns out:
[tex]NO_2+H_2O\rightarrow HNO_3[/tex]
Since two hydrogen atoms are present at the reactants we balance it as follows:
[tex]3NO_2+H_2O\rightarrow 2HNO_3+NO[/tex]
In such a way, since there is a 1:2 mole ratio between water and nitric acid, the produced moles of nitric acid, turns out:
[tex]n_{HNO_3}=0.292molH_2O*\frac{2molHNO_3}{1molH_2O}\\\\ n_{HNO_3}=0.584molHNO_3[/tex]
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