How many moles of $P_{4}$ can be produced by the reaction of $0.10 moles C a_{5} {(P O_{4})}_{3} F, 0.36 moles S i O_{2}, and 0.90 moles C$ according to the following reaction? $C a_{5} (P O_{4})_{3} F + 18 S i O_{2} + 30 C \to 3 P_{4} + 2 C a F_{2} + 18 C a S i O_{3} + 30 C O$

0.06

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0.03

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0.45

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0.75

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Solution

The correct option is A $0.06$
The given balanced equation is,
$C a_{5} (P O_{4})_{3} F + 18 S i O_{2} + 30 C \to 3 P_{4} + 2 C a F_{2} + 18 C a S i O_{3} + 30 C O$
If $0.10 moles C a_{5} {(P O_{4})}_{3} F, 0.36 moles S i O_{2}, and 0.90 moles C$
reacts to produce $P_{4}$ , the limitting reagent here is $S i O_{2}$ .
From the stoichiometry of the balanced equation,
When 18 moles of $S i O_{2}$ reacts with 3 moles of $P_{4}$ are produced
So 0.36 moles $S i O_{2}$ will give $\frac{3}{18} \times 0.36 m o l o f P_{4} = 0.06 m o l o f P_{4}$

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