Question

A metal forms two oxides. The higher oxide contains 80% metal 0.72g of lower oxide gives 0.8g higher oxide when oxidised. Then the ratio of oxygen that combines with the fixed weight of metal in the two oxides will be?

Answer 1

Step 1: 80% metal is true in higher oxide.

Weight of metal = $\frac{80}{100}\times 0.8=0.64gm$

Weight of oxygen = $\frac{20}{100}\times 0.8=0.16gm$

Step 2:

In lower oxidation, weight of metal will be same as that of higher oxidation state (+2)

Weight of oxygen = (72 - 0.64) = 0.08g

So, the ratio of weight of $O_2$ and metals in two oxide is 0.16 : 0.08 = 2 : 1

1000 mL of 0.75 M HCl = 0.75 mol

Step 3:

25 mL of HCl will contain HCl contain HCl = $0.75\times \frac{25}{1000}=0.01875$
2 mol of HCl reacts with 1 mol of $CaC{O}_3$.

So, 0.01875 mol of HCl will react with $\frac{1}{2}\times 0.01875=0.009375mol$

Molar mass of $CaC{O}_3=100g$

Hence, the mass of 0.009375 mol of $CaC{O}_3$ = no. of moles $\times$ molar mass

$=0.009375\times 100$