If 0.5 moles of BaCl 2 are mixed with 0.2 moles of Na 3 PO 4 the maximum number of moles of Ba 3( PO 4)2 that can be formed, is

Question

If 0.5 moles of BaCl 2 are mixed with 0.2 moles of Na 3 PO 4 the maximum number of moles of Ba 3( PO 4)2 that can be formed, is (A) 0.7 (B) 0.5 (C) 0.30 (D) 0.10

Tardigrade Admin · Accepted Answer

3 BaCl 2+2 Na 3 PO 4 → Ba 3( PO 4)2+6 NaCl Limiting reactant is Na 3 PO 4 0.2 mol Na 3 PO 4 will give Ba 3( PO 4)2 =(1/2) × 0.2=0.1 mol

Q. If $0.5$ moles of $B a C l_{2}$ are mixed with $0.2$ 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

0.7

0.5

0.30

0.10

$3 B a C l_{2} + 2 N a_{3} P O_{4} \to B a_{3} (P O_{4})_{2} + 6 N a Cl$

Limiting reactant is $N a_{3} P O_{4}$

$0.2 m o l N a_{3} P O_{4}$ will give $B a_{3} (P O_{4})_{2}$

$= \frac{1}{2} \times 0.2 = 0.1 m o l$

Q. If 0.5 moles of BaCl2​ are mixed with 0.2 moles of Na3​PO4​ the maximum number of moles of Ba3​(PO4​)2​ that can be formed, is

0.7

0.5

0.30

0.10

3BaCl2​+2Na3​PO4​→Ba3​(PO4​)2​+6NaCl Limiting reactant is Na3​PO4​ 0.2molNa3​PO4​ will give Ba3​(PO4​)2​ =21​×0.2=0.1mol

Q. If $0.5$ moles of $B a C l_{2}$ are mixed with $0.2$ 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

$3 B a C l_{2} + 2 N a_{3} P O_{4} \to B a_{3} (P O_{4})_{2} + 6 N a Cl$

Limiting reactant is $N a_{3} P O_{4}$

$0.2 m o l N a_{3} P O_{4}$ will give $B a_{3} (P O_{4})_{2}$

$= \frac{1}{2} \times 0.2 = 0.1 m o l$