How many moles of P_4 can be produced by reaction of 0.10 moles C… - Vidyadip

MCQ

How many moles of $P_4$ can be produced by reaction of $0.10$ moles $Ca_5(PO_4)_3F, 0.36$ moles $SiO_2$ and $0.90$ moles $C$ according to the following reaction?

$4C{a_5}{(P{O_4})_3}F + 18Si{O_2} + 30C \to 3{P_4} + 2Ca{F_2} + 18CaSi{O_3} + 30CO$

✓
$0.060$
B
$0.030$
C
$0.045$
D
$0.075$

✓

Answer

Correct option: A.

$0.060$

a
$\mathrm{SiO}_{2}=$ Limiting reactant

$18 \mathrm{SiO}_{2} \stackrel{\text { give }}{\longrightarrow} 3 \mathrm{P}_{4}$

$0.36 \,\mathrm{mol} \,\mathrm{SiO}_{2} \rightarrow \frac{3}{18} \times 0.36=0.06\, \mathrm{mol}\, \mathrm{P}_{4}$

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