If 0.5 mol of $B a C l_{2}$ is mixed with 0.2 mol of $N a_{3} P O_{4}$ , the maximum number of moles of $B a_{3} (P O_{4})_{2}$ that can be formed is:

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
$3 B a C l_{2} + 2 N a_{2} P O_{4} \to B a_{3} (P O_{4})_{2} + 3 N a C l$
Limiting reagent is $N a_{3} P O_{4}$
0.2 mol $N a_{3} P O_{4}$ will give = $\frac{1}{2} \times 0.2 = 0.1$ mol of $B a_{3} (P O_{4})_{2}$

Suggest Corrections

Similar questions

If $0.5 m o l e$ of $B a C l_{2}$ is mixed with $0.2 m o l e$ of $N a_{3} P O_{4}$ , the maximum number of moles of $B a_{3} (P O_{4})_{2}$ formed are

If 0.5 mole of BaCl₂ is mixed with 0.2 mole of Na₃PO₄,then maximum number of moles of Ba₃(PO₄)₂that can be formed

Q. If

0.5

moles of

B a C l_{2}

is mixed with

0.20

moles of

N a_{3} P O_{4}

, the maximum number of moles of

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

that can be formed is :

Q. 0.5 mole

B a C l_{2}

is mixed with 0.2 mole

N a_{3} P O_{4}

, the maximum number of moles of

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

that can be formed is :

Q. If 0.5 moles of

B a C l_{2}

is mixed with 0.1 moles of

N a_{3} P O_{4},

the maximum amount of

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

that can be formed is

Join BYJU'S Learning Program

If 0.5 mol of BaCl2 is mixed with 0.2 mol of Na3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed is:

The correct option is D 3BaCl2+2Na2PO4→Ba3(PO4)2+3NaCl Limiting reagent is Na3PO4 0.2 mol Na3PO4 will give = 12×0.2=0.1 mol of Ba3(PO4)2

If 0.5 mol of $B a C l_{2}$ is mixed with 0.2 mol of $N a_{3} P O_{4}$ , the maximum number of moles of $B a_{3} (P O_{4})_{2}$ that can be formed is:

The correct option is D
$3 B a C l_{2} + 2 N a_{2} P O_{4} \to B a_{3} (P O_{4})_{2} + 3 N a C l$
Limiting reagent is $N a_{3} P O_{4}$
0.2 mol $N a_{3} P O_{4}$ will give = $\frac{1}{2} \times 0.2 = 0.1$ mol of $B a_{3} (P O_{4})_{2}$