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If the percent yield for the following reaction is 75.0%, and 40.0 g of NO2 are consumed in the reaction, how many grams of nitric acid, HNO3(aq) are produced? 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) If the percent yield for the following reaction is 75.0%, and 40.0 g of NO2 are consumed in the reaction, how many grams of nitric acid, HNO3(aq) are produced? 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g)

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sebassandin

Answer:

[tex]m_{HNO_3}=27.4gHNO_3[/tex]

Explanation:

Hello,

In this case, for the chemical reaction:

[tex]3 NO_2(g) + H_2O(l) \rightarrow 2 HNO_3(aq) + NO(g)[/tex]

The first step is to compute the theoretical yield of nitric acid via stoichiometry in terms of the 3:2 ratio between nitrogen dioxide (molar mass = 46 g/mol) and nitric acid (molar mass = 63 g/mol) respectively:

[tex]m_{HNO_3}^{theoretical}=40.0gNO_2*\frac{1molNO_2}{46gNO_2}* \frac{2molHNO_3}{3molNO_2} *\frac{63gHNO_3}{1molHNO_3} \\\\m_{HNO_3}^{theoretical}=36.52gHNO_3[/tex]

Now, the actual amount is computed by taking into account the 75.0-% percent yield:

[tex]m_{HNO_3}=0.75*36.5gHNO_3\\\\m_{HNO_3}=27.4gHNO_3[/tex]

Best regards.

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