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 \(Cu_2O\)
