Answer:
0.0793 mol.
Explanation:
Refer to a modern periodic table for relative atomic mass data:
Formula mass of copper(II) nitrate, [tex]\rm Cu(NO_3)_2[/tex]:
[tex]\begin{aligned}M({\rm Cu(NO_3)_2})&=\underbrace{63.546}_{\rm Cu} + 2\times(\;\rlap{$\overbrace{\phantom{14.007 + 3\times 15.999}}^{\rm {NO_3}^{-}}$}\underbrace{14.007}_{\rm N} + 3\times\underbrace{15.999}_{\text{O}}\;)\\&=\rm 187.55\; g\cdot mol^{-1}\end{aligned}[/tex].
Number of moles of copper(II) nitrate produced:
[tex]\begin{aligned} n &= \frac{m}{M}\\&= \rm \frac{5.58\; g}{187.55\; g\cdot mol^{-1}} \\&= 0.029751\; mol\end{aligned}[/tex].
The ratio between the coefficient of [tex]\rm HNO_3[/tex] and that of [tex]\rm Cu(NO_3)_2[/tex] in the balanced equation is:
[tex]\displaystyle \frac{n({\rm HNO_3})}{n({\rm Cu(NO_3)_2})} = \frac{8}{3}[/tex].
In other words,
[tex]\begin{aligned} n({\rm HNO_3})&= \frac{8}{3}\cdot n({\rm Cu(NO_3)_2})\\&= \rm \frac{8}{3} \times 0.029751\; mol\\ &=\rm 0.0793\; mol \end{aligned}[/tex].
