If 5 moles of $C a C l_{2}$ is mixed with 2 moles of $N a_{3} P O_{4}$ the maximum number of moles of $C a_{3} (P O_{4})_{2}$ formed will be:
$C a C l_{2} + N a_{3} P O_{4} \to C a_{3} (P O_{4})_{2} + N a C l$

Open in App

Solution

Balanced chemical equation:
$3 C a C l_{2} + 2 N a_{3} P O_{4} \to C a_{3} (P O_{4})_{2} + 6 N a C l$
Finding limiting reagent:
$\frac{moles}{stoichiometric coefficient} = \frac{moles of calcium chloride}{3}$
$= \frac{5}{3} = 1.66$
$\frac{moles of sodium phosphate}{2} = \frac{2}{2} = 1$
Hence sodium phosphate is the limiting reagent.
$\frac{moles of sodium phosphate}{2} = \frac{moles of calcium phosphate}{1}$
Moles of calcium phosphate formed = 1 mol

Suggest Corrections

Similar questions

Q. If 5 moles of

C a C l_{2}

is mixed with 2 moles of

N a_{3} P O_{4}

the maximum number of moles of

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

formed will be:

C a C l_{2} + N a_{3} P O_{4} \to C a_{3} (P O_{4})_{2} + N a C l

Q. If

0.50

mol of

C a C l_{2}

is mixed with

0.20

mol of

N a_{3} P O_{4}

, the maximum number of moles of

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

that can be formed is:

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

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 :

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

Join BYJU'S Learning Program

If 5 moles of CaCl2 is mixed with 2 moles of Na3PO4 the maximum number of moles of Ca3(PO4)2 formed will be: CaCl2+Na3PO4→Ca3(PO4)2+NaCl

If 5 moles of $C a C l_{2}$ is mixed with 2 moles of $N a_{3} P O_{4}$ the maximum number of moles of $C a_{3} (P O_{4})_{2}$ formed will be:
$C a C l_{2} + N a_{3} P O_{4} \to C a_{3} (P O_{4})_{2} + N a C l$