Chemistry - Complete d-block Question with Solution | TestHub

Ferrochrome ore contains $FeC{r}_2{O}_4$.

On fusion with NaOH and excess $NaN{O}_3$, chromium is converted into sodium chromate:

$$4FeC{r}_2{O}_4+16NaOH+7O_2⟶8N{a}_{2}Cr{O}_4+2F{e}_2{O}_3+8H_{2}O$$

After adding water:

Residue $X=F{e}_2{O}_3$

Filtrate $Y=N{a}_{2}Cr{O}_4$ (yellow solution)

Now, when excess $AgN{O}_3$ is added:

$$N{a}_{2}Cr{O}_4+2AgN{O}_3⟶A{g}_{2}Cr{O}_4\downarrow +2NaN{O}_3$$

So precipitate $Z=A{g}_{2}Cr{O}_4$.

When filtrate $Y$ is treated with $H_{2}S{O}_4$, sodium chromate converts into sodium dichromate:

$$2N{a}_{2}Cr{O}_4+H_{2}S{O}_4⟶N{a}_{2}C{r}_2{O}_7+N{a}_{2}S{O}_4+H_{2}O$$

So compound $u=N{a}_{2}C{r}_2{O}_7$.

Molar mass:

$N{a}_{2}C{r}_2{O}_7=(2\times 23)+(2\times 52)+(7\times 16)=46+104+112=262g/mol$