Question

On reduction with hydrogen, 3.6g of an oxide of metal left 3.2g of the metal. If the atomic weight of the metal is 64, the simplest formula of the oxide would be:
A. ${{M}_{2}}{{O}_{2}}$
B. ${{M}_{2}}O$
C. MO
D. $M{{O}_{2}}$

Answer 1

Metal oxide are formed by reacting metal with an oxygen. The metal oxide generally has an oxidation number of -2 and exists as an oxygen anion. During the extraction of metal, the metallic oxides are generally reduced to metals.

Complete answer:
To approach this question let us assume that the number of atoms of the metal M present be x and the number of atoms of oxygen be y, so the metal oxide will be ${{M}_{x}}{{O}_{y}}$
So the reaction of metal oxide with hydrogen will be the following:
${{M}_{x}}{{O}_{y}}+y{{H}_{2}}\to xM+y{{H}_{2}}O$
The moles of metal will be= $\dfrac{given\; weight\; of\; metal}{atomic\; weight\; of\; metal}$
- Given weight of metal =3.2g
- Atomic weight of metal= 64g/mol
On substituting the value in formula we get,
Moles of metal = $\dfrac{3.2}{64}=0.05moles$

Now the moles of metal oxide will be = $\dfrac{given\; weight\; of\; metal\; oxide}{molecular\; weight\; of\; metal\; oxide}$
- Given weight of metal oxide= 3.6
- Molecular weight of metal oxide= 64x+16y
Substituting the value in the formula we get,
Moles of metal oxide = $\dfrac{3.6}{64x+16y}$

As we see in the equation that 1 mole of metal oxide is given x mole of metal M
Then according to this we can write:
$\dfrac{0.05}{x}=\dfrac{3.6}{64x+16y}$
On simplifying we get,

64x+16y=72x \\\ 8x=16y \\\ x=2y $$ So the formula of metal oxide will be $${{M}_{x}}{{O}_{y}}={{M}_{2y}}{{O}_{y}}$$so on simplifying the formula we will get $${{M}_{2}}O$$. **Hence the correct answer is option B.** **Note:** Reduction is defined as losing electrons or the loss of oxygen to form the product. In means of hydrogen reduction is defined as the gain of hydrogen. The word reduction is Latin word meaning ‘to lead back’. In the reaction in which oxidation and reduction occurs simultaneously is called redox reaction.