# If 5.04 grams of iron (II) nitrate reacts with an excess of H_2S, what mass of iron(II) sulfide will be produced in the reaction Fe(NO_3)_2 + H_2S -> FeS + H(NO_3)?

Aug 8, 2016

Mass of ferrous nitrate $=$ $\frac{5.04 \cdot g}{179.86 \cdot g \cdot m o {l}^{-} 1} \times 87.91 \cdot g \cdot m o {l}^{-} 1$ $=$ ??g

#### Explanation:

$F e {\left(N {O}_{3}\right)}_{2} \left(a q\right) + {H}_{2} S \left(a q\right) \rightarrow F e S \left(s\right) \downarrow + 2 H N {O}_{3} \left(a q\right)$

The stoichiometrically balanced equation tells you explicitly that 1 mol of sulfide will result from every mole of ferrous nitrate.

$\text{Moles of ferrous nitrate}$ $=$ $\frac{5.04 \cdot g}{179.86 \cdot g \cdot m o {l}^{-} 1}$ $=$ ??

$\text{Mass of ferrous sulfide}$ $=$ $\frac{5.04 \cdot g}{179.86 \cdot g \cdot m o {l}^{-} 1} \times 87.91 \cdot g \cdot m o {l}^{-} 1$ $\cong$ $2.0 \cdot g$

The sulfide will be obtained as a sticky, black mess that is exceptionally difficult to handle and dry.