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`

Step-by-step 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}=(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* 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}=(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