Question

When 1.187 g of a metallic oxide is reduced with excess hydrogen 1.054 g of the metal is produced. what is the metallic oxide?

Answer 1

Answer: The metallic oxide formed will be $Cu_2O$

Explanation:

We are given a metallic oxide, having general chemical formula $M_2O_n$

We are given:

Mass of metallic oxide = 1.187 g

Mass of metal = 1.054 g

Mass of oxygen = 1.187 - 1.054 = 0.133 g

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .......(1)

Given mass of oxygen = 0.133 g

Molar mass of oxygen = 16 g/mol

Putting values equation 1, we get:

$\text{Moles of oxygen}=\frac{0.133g}{16g/mol}=0.0083mol$

Number of moles of metal in the oxide is twice than the number of moles of oxygen

Number of moles of metal = $(2\times 0.0083)=0.0166$ moles

Now, calculating the molar mass of metal by using equation 1, we get:

Moles of metal = 0.0166 moles

Mass of metal = 1.054 g

Putting values in equation 1, we get:

$0.0166mol=\frac{1.054g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.054g}{0.0166mol}=63.49g/mol$

The metal having 63.49 g/mol as molar mass is copper

Hence, the metallic oxide formed will be $Cu_2O$

Answer 2

Mass O = 1.187 - 1.054 = 0.133 g 
Moles O = 0.133 g / 16 g/mol = 0.00831 
Moles metal = 2 x 0.00831 = 0.0166 
Atomic mass metal = 1.054 g / 0.0166 = 63.4 g/mol 
( this is the atomic mass of Cu) 
so the metal oxide is Cu2O