| How many moles of \[{{P}_{4}}\] can be produced by reaction of 0.10 moles \[C{{a}_{5}}{{(P{{O}_{4}})}_{3}}\text{F},0.36\] moles \[Si{{O}_{2}}\] and 0.90 moles C according to the following reaction? |
| \[4C{{a}_{5}}{{\left( P{{O}_{4}} \right)}_{3}}F+18Si{{O}_{2}}+30C\xrightarrow{{}}\]\[3{{P}_{4}}+2Ca{{F}_{2}}+18CaSi{{O}_{3}}+30CO\] |
A) 0.060
B) 0.030
C) 0.045
D) 0.075
Correct Answer: A
Solution :
[a] \[4C{{a}_{5}}{{\left( P{{O}_{4}} \right)}_{3}}F+18\underset{0.36}{\overset{L.R.}{\mathop{Si}}}\,{{O}_{2}}+\underset{0.9}{\mathop{30C}}\,\xrightarrow{{}}3{{P}_{4}}+2Ca{{F}_{2}}\]\[+18CaSi{{O}_{3}}+30CO\] 18 moles of \[Si{{O}_{2}}\] gives 3 moles of \[{{P}_{4}}\] 0.36 moles of \[Si{{O}_{2}}\]will give \[=\frac{3}{18}\times 0.36=0.06\]mole
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