Question

A metal oxide has the formula Z₂O₃. It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. What is the atomic weight of the metal ?

Answer 1

Writing the concerned equation

\(\underset{0.1596 g}{Z_2O_3}+\underset{6mg = 0.006 g}{3H_2} \longrightarrow 2Z + 3H_2O\)

Thus 0.006 g of H₂ reduces 0.1596 g of Z₂O₃

6 g of H₂ will reduce \(\frac{0.1596}{0.006}\) x 6 g of Z₂O₃ = 159.6 g of Z₂O₃

Molecular weight of Z₂O₃ = 159.6

In other words, 2Z + 48 = 159.6

2Z = 159.6 - 48 = 111.6 

Z = \(\frac{111.6}2\) = 55.8