# What volume of 0.250 M HNO3(aq) is required to completely react with 10.0 g of zinc metal according to the equation: 2 HNO3(aq) + Zn(s) → Zn(NO3)2(aq) + H2(g) ?

## What volume of 0.250 M HNO3(aq) is required to completely react with 10.0 g of zinc metal according to the equation: 2 HNO3(aq) + Zn(s) → Zn(NO3)2(aq) + H2(g) ?

May 23, 2016

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

We need over $1 \cdot L$ of nitric acid.

#### Explanation:

The stoichiometric equation shows a $2 : 1$ equivalence between the acid and zinc respectively.

$\text{Moles of Zinc}$ $=$ $\frac{10.0 \cdot g}{65.38 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.153 \cdot m o l$.

Thus we need a $0.306$ molar quantity of acid.

We have $0.250 \cdot m o l \cdot {L}^{-} 1$ nitric acid available, thus,

$\frac{0.306 \cdot \cancel{m o l}}{0.250 \cdot \cancel{m o l} \cdot \cancel{{L}^{-} 1}} \times 1000 \cdot m L \cdot \cancel{{L}^{-} 1}$ $=$ ??*mL