Question

A metal oxide has the formula ${ Z }_{ 2 }{ O }_{ 3 }$. It can be reduced by hydrogen to give free metal and water. 0.1596g of the metal requires 6 mg of hydrogen for complete reduction. The atomic mass of the metal is:
a.) 27.9 g/mol
b.) 159.6 g/mol
c.) 79.8 g/mol
d.) 55.8 g/mol

Answer 1

Hint: By reading the question we can get enough pieces of information to write the equation for the reduction of metal oxide and forming free metal and water. Now use your knowledge of a mole concept to solve this and find the answer.

Complete step by step answer:

First, we will write the balanced equation of reduction for metal oxide -
$Z_{ 2 }O_{ 3 }\quad +\quad 3H_{ 2 }\quad \rightarrow \quad 2Z\quad +\quad 3H_{ 2 }O$
Here, we can see that 2 moles of the metal require 3 moles of $H_{ 2 }$ for complete reduction.
Moles of $H_{ 2 }$will be equal to the weight given divided by the molecular mass of $H_{ 2 }$.
Moles of $H_{ 2 }$ = 6mg/2 = 3x${ 10 }^{ -3 }$ moles
Now, 3 moles of $H_{ 2 }$ form 2 moles of metal Z
So, 1 mole of $H_{ 2 }$ will form 2/3 moles of metal Z
Hence, 3x10-3 moles of $H_{ 2 }$ will form 3x${ 10 }^{ -3 }$x0.66 moles of metal Z
Now, we have the moles of metal and weight of the metal too. So, we can easily calculate the atomic mass of metal Z -
The atomic mass of metal
= gram weight/moles
= 0.1596 g/3x${ 10 }^{ -3 }$x0.66 moles
=79.8 g/mol

Therefore, the correct answer to this question is option C.

Note: Now you have the value of metal Z, from this you can also calculate the molecular mass of $Z_{ 2 }O_{ 3 }$. You just need to add the atomic masses of each atom. It will be approximately equal to $(79\times 82+16\times 3)$ =208 g/mol.