If 5 moles of $\mathrm{BaCl}_{2}$ is mixed with 2 moles of $\mathrm{Na}_{3} \mathrm{PO}_{4}$, the maximum number of moles of $\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}$ formed is ______ (Nearest integer)

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Some Basic Concepts of Chemistry

If 5 moles of

BaCl_{2}

is mixed with 2 moles of

Na_{3} PO_{4}

, the maximum number of moles of

Ba_{3} (PO_{4})_{2}

formed is ______
(Nearest integer)

Answer: 1

Solution

The precipitation follows

3 BaC l_{2} + 2 N a_{3} P O_{4} \to B a_{3} (P O_{4})_{2} + 6 NaCl

. The 5 mol of barium chloride would need

\frac{5 \times 2}{3} = 3.33

mol of sodium phosphate, but only 2 mol is available, so sodium phosphate is the limiting reagent. Since 2 mol of sodium phosphate yields 1 mol of product, the maximum amount of

B a_{3} (P O_{4})_{2}

formed is 1 mol.

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Integer answer

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