How many moles of P₄ can be produced by reaction of 0.10 moles Ca₅(PO₄)₃F,0.36 moles SiO₂, and 0.90 moles C according to the following reaction ?
$4{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}+18{\mathrm{SiO}}_2+30\mathrm{C}⟶3{\mathrm{P}}_4+2{\mathrm{CaF}}_2+18{\mathrm{CaSiO}}_3+30\mathrm{CO}$

$4{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}+\underset{0.36}{\overset{\mathrm{L}\cdot \mathrm{R}}{18{\mathrm{SiO}}_2}}+\underset{0.9}{30\mathrm{C}}⟶3{\mathrm{P}}_4+2{\mathrm{CaF}}_2+18{\mathrm{CaSiO}}_3+30\mathrm{CO}$

SiO₂ is the Limiting reagent ($\because$0.36 moles of SiO₂ requires 0.6moles of Carbon)
18 mole SiO₂ gives 3 mole P₄
0.36 mole SiO₂ will give $=\frac{3}{18}\times 0.36=0.06 \mathrm{mole}$