Question

zinc is determined by precipitating and weighing as zn2fe(cn)6 . what weight of zinc is contained in a sample that gives 0.348 g precipitate? (zn=65, fe=56, c=12, n = 14

Answer 1

To determine the weight of zinc in the sample that produces 0.348 g of Zn2Fe(CN)6 precipitate, we need to follow these steps:

1. **Calculate the molar mass of Zn2Fe(CN)6:**

- Zinc: \( Zn \), there are 2 zinc atoms.
\[ 2 \times 65 = 130 \, \text{g/mol} \]
- Iron: \( Fe \), there is 1 iron atom.
\[ 1 \times 56 = 56 \, \text{g/mol} \]
- Carbon: \( C \), there are 6 carbon atoms.
\[ 6 \times 12 = 72 \, \text{g/mol} \]
- Nitrogen: \( N \), there are 6 nitrogen atoms.
\[ 6 \times 14 = 84 \, \text{g/mol} \]

Adding up all these contributions:
\[ 130 \, (\text{Zn}) + 56 \, (\text{Fe}) + 72 \, (\text{C}) + 84 \, (\text{N}) = 342 \, \text{g/mol} \]

2. **Determine the proportion of zinc in the molar mass of Zn2Fe(CN)6:**

The total molar mass of Zn2Fe(CN)6 is 342 g/mol.
The molar mass contributed by zinc is 130 g/mol.

The fraction of the mass attributed to zinc in Zn2Fe(CN)6 is:
\[ \frac{130}{342} \]

3. **Calculate the weight of zinc in the given precipitate:**

The given weight of the precipitate is 0.348 g.

The weight fraction of zinc in the precipitate is:
\[ 0.348 \, \text{g} \times \frac{130}{342} \]

\[ \text{Weight of zinc} = 0.348 \times \frac{130}{342} \]

\[ \text{Weight of zinc} \approx 0.348 \times 0.38 \]

\[ \text{Weight of zinc} \approx 0.132 \, \text{g} \]

So, the weight of zinc contained in the sample is approximately 0.132 grams.