Answer:
[tex]\large \boxed{\text{106 L}}[/tex]
Explanation:
We will need a balanced chemical equation with molar masses and volumes, so, let's gather all the information in one place.
MV/L: 22.71
M_r: 63.55
Cu + 4HNO₃ ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂
m/g: 149
(a) Moles of Cu
[tex]\text{Moles of Cu } =\text{149 g Cu } \times \dfrac{\text{1 mol Cu }}{\text{63.55 g Cu }} =\text{2.344 mol Cu}[/tex]
(b) Moles of NO₂
The molar ratio is 2 mol NO₂:1 mol Cu
[tex]\text{Moles of NO$_{2}$}= \text{2.344 mol Cu} \times \dfrac{\text{2 mol NO$_{2}$}}{ \text{1 mol Cu}} = \text{4.689 mol NO$_{2}$}[/tex]
(c) Volume of NO₂
The volume of 1 mol of an ideal gas at STP (0 °C and 1 bar) is 22.71 L.
[tex]\text{V} = \text{4.689 mol} \times \dfrac{\text{22.71 L}}{\text{1 mol}} = \textbf{106 L}\\\\\text{You can produce $\large \boxed{\textbf{106 L}} $ of NO$_{2}$.}[/tex]
