JEE MAIN - Chemistry (2024 - 8th April Evening Shift - No. 20)
In qualitative test for identification of presence of phosphorous, the compound is heated with an oxidising agent. Which is further treated with nitric acid and ammonium molybdate respectively. The yellow coloured precipitate obtained is :
$$\mathrm{Na}_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3$$
$$\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12\left(\mathrm{NH}_4\right)_2 \mathrm{MoO}_4$$
$$\mathrm{MoPO}_4 \cdot 21 \mathrm{NH}_4 \mathrm{NO}_3$$
$$\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3$$
Explanation
For test of phosphorus:
$$\mathrm{{H_3}P{O_4} + 12{(N{H_4})_2}Mo{O_4} + 21HN{O_3} \to \mathop {{{(N{H_4})}_3}\,.\,12Mo{O_3}}\limits_{(Yellow\,ppt)} + 21N{H_4}N{O_3} + 12{H_2}O}$$
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