How many moles of $\mathrm{P}_{4}$ can be produced by reaction of $0.10$ moles…
$4 \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}ight)_{3} \mathrm{~F}+18 \mathrm{SiO}_{2}+30 \mathrm{C} \longrightarrow$
$3 \mathrm{P}_{4}+2 \mathrm{CaF}_{2}+18 \mathrm{CaSiO}_{3}+30 \mathrm{CO}$
- $0.060$
- $0.030$
- $0.045$
- $0.075$
Solution
& 4 \mathrm{Ca}_5 \underset{0.1}{\left(\mathrm{PO}_{4}ight)_3} \mathrm{F}+ \stackrel{\mathrm{L.R.}}{\underset{0.36}{18 \mathrm{SiO}_2}}+\underset{0.9}{30 \mathrm{C}} \longrightarrow 3 \mathrm{P}_4+2 \mathrm{CaF}_2 +18 \mathrm{CaSiO}_3+30 \mathrm{CO}
\end{aligned}\)
18 moles of $\mathrm{SiO}_{2}$ gives 3 moles of $\mathrm{P}_{4}$
$0.36$ moles of $\mathrm{SiO}_{2}$ will give
$=\frac{3}{18} \times 0.36=0.06 \mathrm{~mole}$
Asked in: JEE-TOPICTESTS-CHEMISTRY
Practice more SOME BASIC CONCEPTS OF CHEMISTRY questions on Aicharya