Question

10.4 grams of chromium(Cr) reacted with fluorine gas (F2) to produce 25.6 grams of chromium Florine salt reaction : __Cr(s) +__ F2 (g) —> __Cr? F? (S)

1) how many grams reacted?
2) how many miles of flurorine atoms reacted?
3) how many miles of chromium atoms reacted?
4) what is the chemical formula produced?
6) what is the anion
7) what is the cation of this salt?
8) which element went through reduction?

Answer 1

10.4 grams of chromium (Cr) reacted with fluorine gas (F₂) to produce 25.6 grams of chromium

Florine salt reaction: _Cr(s) + _F₂(g) → _Cr? F? (s)

1) how many grams reacted?

2) how many miles of fluorine atoms reacted?

3) how many miles of chromium atoms reacted?

4) what is the chemical formula produced?

6) what is the anion

7) what is the cation of this salt?

8) which element went through reduction?

Solution.

$$\begin{array}{l} \mathrm{Cr}(s) + 2\mathrm{F}_2(g) \rightarrow \mathrm{CrF}_4 \\ n(\mathrm{Cr}) = \frac{m(\mathrm{Cr})}{M(\mathrm{Cr})} = \frac{10.4}{52} = 0.2 \text{ mol} \\ n(\mathrm{Cr}) = \frac{m(\mathrm{CrF}_4)}{M(\mathrm{CrF})} = \frac{25.6}{128} = 0.2 \text{ mol} \\ n(\mathrm{F}_2) = n(\mathrm{Cr}) \times 2 = 0.2 \times 2 = 0.4 \text{ mol} \\ m(\mathrm{F}_2) = n(\mathrm{F}_2) \times M(\mathrm{F}_2) = 0.4 \times 38 = 15.2 \text{ g} \\ \end{array}$$

The anion is F⁻.

The cation is Cr⁴⁺.

The element F went through reduction.

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