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?

The metallic oxide formed will be

Explanation:

We are given a metallic oxide, having general chemical formula

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:

    .......(1)

Given mass of oxygen = 0.133 g

Molar mass of oxygen = 16 g/mol

Putting values equation 1, we get:

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

Number of moles of metal = 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:

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

Hence, the metallic oxide formed will be