Question

A metal oxide has the formula $M_2O_3$. It can be reduced by $H_2$ to give free metal and water. $0.1596\, g$ of $M_2O_3$ required 6 mg of $H_2$ for complete reduction. The atomic mass of the metal is

• A. $27.9$
• B. $79.8$
• C. $55.8$
• D. $159.8$

Answer 1

The Correct Option is C

To solve this problem, we start by considering the chemical reaction for the reduction of the metal oxide M_2O_3 with hydrogen:

M_2O_3 + 3H_2 \rightarrow 2M + 3H_2O

In this reaction, 3 moles of hydrogen gas (H_2) are needed to completely reduce 1 mole of M_2O_3 to 2 moles of the metal (M). Let's calculate the molar masses involved:

1. The mass of H_2 used is 6 mg which equals 0.006 g.
2. The molar mass of H_2 is 2 g/mol.
3. Calculate the moles of H_2:
   \text{Moles of } H_2 = \frac{0.006\, \text{g}}{2\, \text{g/mol}} = 0.003 \, \text{moles}

According to the balanced equation, 3 moles of H_2 reduce 1 mole of M_2O_3. Therefore, the moles of M_2O_3 are:

\text{Moles of } M_2O_3 = \frac{0.003}{3} = 0.001 \, \text{moles}

We are given that 0.1596 g of M_2O_3 was reduced:

\text{Molar mass of } M_2O_3 = \frac{0.1596 \, \text{g}}{0.001 \, \text{moles}} = 159.6 \, \text{g/mol}

Let's express the molar mass of M_2O_3 in terms of the atomic mass of M (let's call it x):

\text{Molar mass of } M_2O_3 = 2x + 3 \times 16 \, \text{(oxygen's molar mass)} = 2x + 48

Setting the two expressions for the molar mass of M_2O_3 equal:

2x + 48 = 159.6

2x = 159.6 - 48 = 111.6

x = \frac{111.6}{2} = 55.8

Thus, the atomic mass of the metal M is 55.8 g/mol.

Therefore, the correct answer is 55.8.