JEE Main 2023 — Some Basic Concepts Of Chemistry Question with Solution

Given the balanced chemical equation:

$3{\mathrm{BaCl}}_2+2{\mathrm{Na}}_3{\mathrm{PO}}_4\to {\mathrm{Ba}}_3{(\mathrm{PO}}_4{)}_2+6\mathrm{NaCl}$

We can see that 3 moles of ${\mathrm{BaCl}}_2$ react with 2 moles of ${\mathrm{Na}}_3{\mathrm{PO}}_4$ to produce 1 mole of ${\mathrm{Ba}}_3{(\mathrm{PO}}_4{)}_2$.

Given that you have 5 moles of ${\mathrm{BaCl}}_2$ and 2 moles of ${\mathrm{Na}}_3{\mathrm{PO}}_4$, let's calculate the maximum number of moles of ${\mathrm{Ba}}_3{(\mathrm{PO}}_4{)}_2$ that can be formed:

From ${\mathrm{BaCl}}_2$, using the molar ratio from the balanced equation:

$5\text{ moles }{\mathrm{BaCl}}_2\times \frac{1\text{ mole }{\mathrm{Ba}}_3{(\mathrm{PO}}_4{)}_2}{3\text{ moles }{\mathrm{BaCl}}_2}=\frac{5}{3}\text{ moles }{\mathrm{Ba}}_3{(\mathrm{PO}}_4{)}_2$

From ${\mathrm{Na}}_3{\mathrm{PO}}_4$, using the molar ratio from the balanced equation:

$2\text{ moles }{\mathrm{Na}}_3{\mathrm{PO}}_4\times \frac{1\text{ mole }{\mathrm{Ba}}_3{(\mathrm{PO}}_4{)}_2}{2\text{ moles }{\mathrm{Na}}_3{\mathrm{PO}}_4}=1\text{ mole }{\mathrm{Ba}}_3{(\mathrm{PO}}_4{)}_2$

The limiting reactant is ${\mathrm{Na}}_3{\mathrm{PO}}_4$ because it produces 1 mole of ${\mathrm{Ba}}_3{(\mathrm{PO}}_4{)}_2$, which is less than the $\frac{5}{3}$ moles produced by ${\mathrm{BaCl}}_2$.

Therefore, the maximum number of moles of ${\mathrm{Ba}}_3{(\mathrm{PO}}_4{)}_2$ formed is 1 mole. Rounding to the nearest integer, the answer is 1.