Question #8e3d3

Aug 3, 2017

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

Explanation:

We have $\frac{2.00 \cdot g}{65.4 \cdot g \cdot m o {l}^{-} 1} = 0.0306 \cdot m o l$ $\text{zinc metal}$.

And $\frac{2.50 \cdot g}{169.87 \cdot g \cdot m o {l}^{-} 1} = 0.0147 \cdot m o l$ $\text{silver nitrate}$.

And so the zinc metal reductant is slightly in stoichiometric excess.

We form $2 \times 0.0147 \cdot m o l \times 107.9 \cdot g \cdot m o {l}^{-} 1 = 3.17 \cdot g$ with respect to silver metal.

I will let you answer the last question. There are $1.2 \cdot m m o l$ of EXCESS zinc metal..