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.

Answer 1

To calculate the number of moles of water produced in the reaction given, we need to analyze 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 the stoichiometric relationships between the reactants and products. Specifically, the coefficients indicate that for every 3 moles of \(\text{Cu(NO}_3)_2\) produced, 4 moles of \(\text{H}_2\text{O}\) are produced.

Now, since we have 3.3 moles of \(\text{Cu(NO}_3)_2\) produced, we can set up a proportion to find the number of moles of \(\text{H}_2\text{O}\) produced:

\[ \text{Moles of } \text{H}_2\text{O} = \left(\frac{4 \text{ moles } \text{H}_2\text{O}}{3 \text{ moles } \text{Cu(NO}_3)_2}\right) \times 3.3 \text{ moles } \text{Cu(NO}_3)_2 \]

Calculating this gives:

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

Thus, the number of moles of water produced is \(4.4\) moles.