Answer to Question #177648 in General Chemistry for kyle

Question

A mixture of Mg and Zn with a combined mass of 1.0875 g was burned in oxygen producing MgO and ZnO with a combined mass of 1.4090 g. How many grams of zinc were in original mixture?

Answer 1

2Zn + O₂ → 2ZnO

2Mg + O₂ → 2MgO

Let z be the mass of zinc to be found.

Then (1.0875 − z) is the mass of Mg in the original mixture.

Zinc:

$\frac{z}{65.3820 \;g/mol} \times \frac{2 \;mol}{2 \;mol} \times \frac{40.3044 \;g}{1 \; mol} = 1.80337 -1.65827z$

In other words, the amount of magnesium oxide produced is 1.80337 minus 1.65827 times the mass of magnesium in the original mixture.

Add the two masses of oxides and set the sum equal to the given total mass:

(1.244676z) + (1.80337 − 1.65827z) = 1.4090 g

z = 0.9535 g (of zinc)

Answer: 0.9535 g