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 was in original mixture

Answer 1

0.9537g is the original mass of Zn in the mixture

Step-by-step explanation:

We can solve this question using the molar mass of each compound involved:

Mg = 24.305g/mol

MgO = 40.304g/mol

Zn = 65.38g/mol

ZnO = 81.38g/mol

Mg-Zn:

1.0875g = 24.305X + 65.38Y (1)

Where X are moles of Mg and Y moles of Zn

MgO-ZnO:

1.4090g = 40.304X + 81.38Y (2)

(1.4090g - 81.38Y) / 40.304 = X

0.03496 - 2.0192Y = X

Replacing (2) in (1)

1.0875g = 24.305(0.03496 - 2.0192Y) + 65.38Y

1.0875 = 0.8497 - 49.077Y + 65.38Y

0.2378 = 16.303Y

0.01459 moles of Zn = Y

And the mass is:

0.01459 moles of Zn * (65.38g/mol) =

0.9537g is the original mass of Zn in the mixture