Question

Calculate the number of moles of water produced when 3.3 moles of Cu(NO3)2 are formed in the reaction 3Cu + 8HNO3 --> 3Cu(NO3)2 + 2NO + 4H2O.
Responses

0.3 mol
0.3 mol

2.2 mol
2.2 mol

4.4 mol
4.4 mol

19.8 mol
19.8 mol

Answer 1

To determine the number of moles of water produced when 3.3 moles of Cu(NO3)2 are formed, we can refer to the stoichiometry of the balanced chemical equation:

\[ 3 \, \text{Cu} + 8 \, \text{HNO}_3 \rightarrow 3 \, \text{Cu(NO}_3)_2 + 2 \, \text{NO} + 4 \, \text{H}_2\text{O} \]

From the equation, we can see that for every 3 moles of Cu(NO3)2 produced, 4 moles of H2O are produced.

Now we can set up a proportion using the stoichiometric coefficients:

\[
\text{Moles of H}_2\text{O} = \left( \frac{4 \, \text{moles H}_2\text{O}}{3 \, \text{moles Cu(NO}_3)_2} \right) \times \text{moles of Cu(NO}_3)_2
\]

Now substitute 3.3 moles of Cu(NO3)2 into this equation:

\[
\text{Moles of H}_2\text{O} = \left( \frac{4}{3} \right) \times 3.3
\]

Calculating this:

\[
\text{Moles of H}_2\text{O} = \frac{4 \times 3.3}{3} = \frac{13.2}{3} = 4.4 \, \text{moles of H}_2\text{O}
\]

Thus, when 3.3 moles of Cu(NO3)2 are produced, 4.4 moles of water (H2O) are produced.

The response is:
**4.4 mol**