The maximum number of moles of mathrmBa3( mathrmPO4)2 that can be formed if 2mole BaCl2 is mixed with 1mole Na3PO4 is

Question

The maximum number of moles of mathrmBa3( mathrmPO4)2 that can be formed if 2mole BaCl2 is mixed with 1mole Na3PO4 is (A) 0.7 (B) 0.25 (C) 0.33 (D) 0.50

Tardigrade Admin · Accepted Answer

The balanced chemical reaction is 3 mathrmBaCl2+2 mathrmNa3 mathrmPO4 arrow mathrmBa3( mathrmPO4)2+6 mathrmNaCl According to the reaction, 3 mole mathrmBaCl2 reacts with 2 mole mathrmNa3( mathrmPO4)2 Thus, 2moleBaCl2 will require (2/3)× 2=(4/3)=1.33moleNa3PO4 Since sodium phosphate is taken in less quantity then the required quantity, here, Na3PO4 is the limiting reactant. The stoichiometric calculations are done based on the limiting reagent. According to the reaction. 2 mole Na 3 PO 4 gives 1 mole Ba 3( PO 4)2 So, 1 mole Na 3 PO 4 will give 0.5 mole Ba 3( PO 4)2.

Q. The maximum number of moles of $Ba_{3} (PO_{4})_{2}$ that can be formed if $2 m o l e$ $B a C l_{2}$ is mixed with $1 m o l e$ $N a_{3} P O_{4}$ is

$0.7$

$0.25$

$0.33$

$0.50$

Q. The maximum number of moles of Ba3​(PO4​)2​ that can be formed if 2mole BaCl2​ is mixed with 1mole Na3​PO4​ is

0.7

0.25

0.33

0.50

Q. The maximum number of moles of $Ba_{3} (PO_{4})_{2}$ that can be formed if $2 m o l e$ $B a C l_{2}$ is mixed with $1 m o l e$ $N a_{3} P O_{4}$ is

$0.7$

$0.25$

$0.33$

$0.50$