Question

A mixture of powdered aluminum and tin was burned in an atmosphere of oxygen in a way such that the resulting oxides could be collected and weighed 0.5488g; the mixture of Al2O3 and SnO2 weighed 0.7712g. Calculate the weight and percent of Al and Sn in the initial mixture

Answer 1

Answer on Question #81671, Chemistry / General Chemistry

A mixture of powdered aluminum and tin was burned in an atmosphere of oxygen in a way such that the resulting oxides could be collected and weighed 0.5488g; the mixture of $\mathrm{Al}_{2}\mathrm{O}_{3}$ and $\mathrm{SnO}_2$ weighed 0.7712g. Calculate the weight and percent of Al and Sn in the initial mixture.

Solution:

Weight of Sn is X and weight of Al is (0.5488-x)

y

$$\begin{array}{l} \mathrm{Sn} + \mathrm{O}_2 = \mathrm{SnO}_2 \\ \mathrm{M} (\mathrm{SnO}_2) = 150.67 \mathrm{~g/mol} \\ \mathrm{M} (\mathrm{Sn}) = 118.69 \mathrm{~g/mol} \\ y = x * 150.67 / 118.69 = 1.2694x \\ \end{array}$$

z

$$\begin{array}{l} 4 \mathrm{Al} + 3 \mathrm{O}_2 = 2 \mathrm{Al}_2\mathrm{O}_3 \\ \mathrm{M} (\mathrm{Al}) = 4 * 26.98 \\ \mathrm{M} (\mathrm{Al}_2\mathrm{O}_3) = 2 * 101.93 \\ \end{array}$$ $$z = \frac{(0.5488 - x) * 2 * 101.93}{4 * 26.98} = \frac{(0.5488 - x) * 203.86}{107.92} = 1.0366 - 1.8889x$$ $$y + z = 0.7712$$ $$\begin{array}{l} 1.2694x + 1.0366 - 1.8889x = 0.7712 \\ 0.6195x = 0.2654 \\ x = 0.4284 \\ \end{array}$$

Weight of Sn is 0.4284g

$$(0.5488 - 0.4284) = 0.1204$$

Weight of Al is 0.1204g