JEE Main 2025 — D And F Block Elements Question with Solution

$$\frac{Fe{C}_2{O}_4+N{a}_{2}C{O}_3\text{buildrel}{O}_2}{⟶?}$$

The chemical reaction can be written as

The insoluble product is $F{e}_2{O}_3$.

Molar mass of $F{e}_2{O}_3$

Fe (Atomic mass) $\to$ 55.845 amu

Molar mass = 55.845 g mol${}^{-1}$

O (Atomic mass) $\to$ 15.999 amu

Molar mass = 15.999 g mol${}^{-1}$

Molar mass of $F{e}_2{O}_3$ = 55.845 g mol${}^{-1}\times 2$ + 15.999 g mol${}^{-1}\times 3$

$=(111.69+47.997)$ g mol${}^{-1}$

$=159.687$ g mol${}^{-1}\approx 160$ g mol${}^{-1}$