What volume of carbon dioxide would evolve if....?

What volume of carbon dioxide would evolve if a $25 \cdot g$ mass of zinc carbonate is treated with a $0.300 \cdot {\mathrm{dm}}^{3}$ volume of $2 \cdot m o l \cdot {\mathrm{dm}}^{-} 3$ nitric acid?

Apr 15, 2017

We interrogate the following reaction:

$Z n C {O}_{3} \left(s\right) + 2 H N {O}_{3} \left(a q\right) \rightarrow Z n {\left(N {O}_{3}\right)}_{2} \left(a q\right) + {H}_{2} O \left(l\right) + C {O}_{2} \left(g\right) \uparrow$

Explanation:

$\text{Moles of zinc carbonate}$ $=$ $\frac{25.0 \cdot g}{125.42 \cdot g \cdot m o {l}^{-} 1} = 0.199 \cdot m o l .$

$\text{Moles of nitric acid}$ $=$ $0.300 \cdot {\mathrm{dm}}^{3} \times 2 \cdot m o l \cdot {\mathrm{dm}}^{-} 3 = 0.60 \cdot m o l .$

clearly there is excess acid, and the reaction is stoichiometric in zinc carbonate. We should get $0.199 \cdot m o l$ of $C {O}_{2} \left(g\right)$ evolved. Under standard conditions (and you should check which standard conditions pertain - they vary across each syllabus), the molar volume is $22.7 \cdot L \cdot m o {l}^{-} 1$. And thus.........

$22.7 \cdot L \cdot m o {l}^{-} 1 \times 0.60 \cdot m o l = 13 - 14 \cdot L$.