Question

In an experiment, 4g of ${M_2}{O_X}$ oxide was reduced to 2.8 g of the metal. Calculate the number of O atoms in the oxide. (Given atomic mass of the metal= $56gmo{l^{ - 1}}$)

Answer 1

Hint: To find out the number of atoms of oxygen, first we need to determine the moles of metal as mass of metal and atomic number is given. In this reaction reduction of metal oxide to metal is taking place with the elimination of oxygen.

Complete step by step answer:
$2{M_2}{O_X} \to 2 \times 2M + x{O_2}$
Given,
Mass of ${M_2}{O_X}$ is 4g.
The mass of metal is 2.8 g.
The atomic mass of metal is 56 g/mol.
The formula for calculating the moles is shown below.
$n = \dfrac{m}{M}......(i)$
Where
n is the number of moles of the compound.
m is the mass of the compound.
M is the molar mass of the compound.
To calculate the moles of metal substitute the value of mass and atomic mass in equation (i).
$n = \dfrac{{2.8g}}{{56g/mol}}$
$\Rightarrow n = 0.05mol$
From the reaction, it is clear that 2 moles of metal oxide are equivalent to 4 moles of metal and x moles of oxygen.
Moles of ${M_2}{O_X} = \dfrac{2}{4} \times 0.05$
$\Rightarrow n = \dfrac{{0.1}}{4}mol$
To calculate the molecular weight of ${M_2}{O_X}$, substitute the values in equation (i).
\[{M_{{M_2}{O_X}}} = \dfrac{4}{{\dfrac{{0.1}}{4}}}\]
$\Rightarrow (2 \times 56) + x(16) = 1.6$
$\Rightarrow x = 3$

Thus, the number of oxygen atoms in metal oxide is 3.

Note:
Make sure to quarter balance the equation that means the number of moles of atoms present in the compound on the left side of the reaction should be equal to the moles of compounds present on the right side of the reaction. The reaction should follow the law of conservation of mass.