Question

A mixture of 2 metals is analyzed by reacting it with dioxygen to completely convert the metals to their oxides. When 3.1g of a Zn/Fe mixture is reacted it produces 4.274g of a mixture of Fe2SO3 and ZnO. What is the mass percent for each metal in the original mixture, show units and steps to final answer.

Answer 1

Answer on Question #48269 – Chemistry – Inorganic Chemistry

Data:

$$\begin{array}{l} (x) g \\ 4 \text{ Fe} \quad + \quad 3 \mathrm{O}_2 = 2 \mathrm{Fe}_2\mathrm{O}_3 \\ 4.56 \text{ g/mol} \quad 2.160 \text{ g/mol} \\ (3.1 - x) g \\ 2 \text{ Zn} \quad + \mathrm{O}_2 = 2 \text{ZnO} \\ 2.65 \text{ g/mol} \quad 2.81 \text{ g/mol} \end{array}$$

Lets express the mass of Fe as $(x) g$ and the mass of Zn as $(3,1 - x) g$. The same situation with the mass of oxides. According to the equations calculate the mass of Fe and Zn using algebraic proportions:

$$\left\{ \begin{array}{l} \frac{x}{4 \cdot 56} = \frac{y}{2 \cdot 160}; \\ \frac{3.1 - x}{2 \cdot 65} = \frac{4.274 - y}{2 \cdot 81}; \end{array} \right.$$ $$\begin{array}{l} 320 x = 224 y; \\ 502.2 - 162 x = 555.62 - 130 y; \end{array}$$ $$\begin{array}{l} x = \frac{224 y}{320}; \\ -53.42 - 162 \frac{224 y}{320} + 130 y = 0; \end{array}$$ $$\begin{array}{l} x = \frac{224 y}{320} \\ - \frac{36288 y}{320} + 130 y = 53.42; \\ x = \frac{224 y}{320}; \\ -113.4 y + 130 y = 53.42; \\ x = \frac{224 y}{320}; \\ 16.6 y = 53.42; \\ x = 2.247; \\ y = 3.21. \end{array}$$

Taking into account calculations the mass of Fe in mixture is $2,247 \text{ g}$, so the mass of Zn is equivalent $0,853 \text{ g}$. The mass percent of both metals is calculated according to formula:

$$W (Me) = \frac{m (Me)}{m (Mixture)} \times 100\%$$ $$W (Fe) = \frac{2.2347 \text{ g}}{3.1 \text{ g}} \cdot 100\% = 72.48\%;$$ $$W (Zn) = \frac{0.853 \text{ g}}{3.1 \text{ g}} \cdot 100\% = 27.52\%.$$

Answer: mass percent for each metal in the original mixture is $72.48\%$ (Fe) and $27.52\%$ (Zn).

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