Question

3 Cu(s) + 8 HNO3(aq) + 3 Cu(NO3)2(aq) + 2NO(9) + 4H20() c) If 5.58 g of copper(II) nitrate, Cu(NO3)2, is eventually obtained, how many moles of nitric acid, HNO3, were used in the experiment?

Answer 1

0.0793 mol.

Step-by-step explanation:

Refer to a modern periodic table for relative atomic mass data:

• Cu: 63.546;
• N: 14.007;
• O: 15.999.

Formula mass of copper(II) nitrate,
`\rm Cu(NO_3)_2`:

`\begin{aligned}M({\rm Cu(NO_3)_2})&=\underbrace{63.546}_(\rm Cu) + 2*(\;\rlap{$\overbrace{\phantom{14.007 + 3* 15.999}}^{\rm {NO_3}^(-)}$}\underbrace{14.007}_(\rm N) + 3*\underbrace{15.999}_{\text{O}}\;)\\&=\rm 187.55\; g\cdot mol^(-1)\end{aligned}`.

Number of moles of copper(II) nitrate produced:

`\begin{aligned} n &= (m)/(M)\\&= \rm (5.58\; g)/(187.55\; g\cdot mol^(-1)) \\&= 0.029751\; mol\end{aligned}`.

The ratio between the coefficient of
`\rm HNO_3` and that of
`\rm Cu(NO_3)_2` in the balanced equation is:

`\displaystyle \frac{n({\rm HNO_3})}{n({\rm Cu(NO_3)_2})} = (8)/(3)`.

In other words,

`\begin{aligned} n({\rm HNO_3})&= (8)/(3)\cdot n({\rm Cu(NO_3)_2})\\&= \rm (8)/(3) * 0.029751\; mol\\ &=\rm 0.0793\; mol \end{aligned}`.