Limiting Reactant Calculation in Precipitation Reactions

If 5moles of BaCl_2 is mixed with 2moles of Na_3PO_4, the maximum number of moles of Ba_3(PO_4)_2 formed is............. (Nearest integer)

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Type the value - units or words beside it are fine.

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why is

{N a}_{3} {P O}_{4}

the limiting reagent when there are fewer moles of it added?

The limiting reagent is determined by the ratio of moles to stoichiometric coefficient ( $e x t m o l e s / coeff$ ), not simply the smallest number of moles. Here, $\frac{2}{2} = 1 < \frac{5}{3} \approx 1.67$ , so ${N a}_{3} {P O}_{4}$ runs out first.

Answer

Accepted answer: 1

The maximum number of moles of ${B a}_{3} ({P O}_{4})_{2}$ formed is $1$ .

Why each option works or fails

Step-by-step solution

Given: Initial amount of ${B a C l}_{2} = 5 moles$ , Initial amount of ${N a}_{3} {P O}_{4} = 2 moles$ .
Find: Maximum number of moles of ${B a}_{3} ({P O}_{4})_{2}$ formed.
Strategise: Write the balanced chemical reaction: $3 {B a C l}_{2} + 2 {N a}_{3} {P O}_{4} \to {B a}_{3} ({P O}_{4})_{2} + 6 N a C l$ Compare the mole-to-coefficient ratios $\frac{n_{i}}{coefficient}$ to find the limiting reagent (LR), then calculate the moles of precipitate formed from the LR.
Execute: For ${B a C l}_{2}$ : $\frac{5}{3} \approx 1.67$ . For ${N a}_{3} {P O}_{4}$ : $\frac{2}{2} = 1.0$ . Since $1.0 < 1.67$ , ${N a}_{3} {P O}_{4}$ is the limiting reagent. From stoichiometry, $2 moles$ of ${N a}_{3} {P O}_{4}$ produce $1 mole$ of ${B a}_{3} ({P O}_{4})_{2}$ . Therefore, moles of ${B a}_{3} ({P O}_{4})_{2}$ formed $= 2 \times \frac{1}{2} = 1$ .
Verify: $2 moles$ of ${N a}_{3} {P O}_{4}$ require $3 moles$ of ${B a C l}_{2}$ . Since we have $5 moles$ of ${B a C l}_{2}$ , $2 moles$ remain unreacted. Thus, exactly $1 mole$ of ${B a}_{3} ({P O}_{4})_{2}$ precipitates.

Limiting Reactant Calculation in Precipitation Reactions

What feels right?

The maximum number of moles of ${B a}_{3} ({P O}_{4})_{2}$ formed is $1$ .

Why each option works or fails

Solution

01Given

02Find

03Strategise

04Execute

✓Verify

We think you should solve this next.

Iron oxide $FeO$ , crystallises in a cubic lattice with a unit cell edge length of…

The graph of $\log \frac{x}{m}$ vs $\log p$ for an adsorption process…

In ammonium-phosphomolybdate, the oxidation state of $M o$ is…

Students also ask

Limiting Reactant Calculation in Precipitation Reactions

The maximum number of moles of Ba3(PO4)2Ba3​(PO4​)2​ formed is 11.

01Given

02Find

03Strategise

04Execute

✓Verify

Iron oxide FeOFeO, crystallises in a cubic lattice with a unit cell edge length of…

The graph of log⁡xmlogmx​ vs log⁡plogp for an adsorption process…

In ammonium-phosphomolybdate, the oxidation state of MoMo is…

Students also ask

The maximum number of moles of ${B a}_{3} ({P O}_{4})_{2}$ formed is $1$ .

Iron oxide $FeO$ , crystallises in a cubic lattice with a unit cell edge length of…

The graph of $\log \frac{x}{m}$ vs $\log p$ for an adsorption process…

In ammonium-phosphomolybdate, the oxidation state of $M o$ is…