A) 0.70
B) 0.50
C) 0.20
D) 0.10
Correct Answer: D
Solution :
\[3\,CaC{{l}_{2}}+2\,N{{a}_{3}}P{{O}_{4}}\to C{{a}_{3}}{{(P{{O}_{4}})}_{2}}+6NaCl\] \[\therefore \] Mole of \[N{{a}_{3}}P{{O}_{4}}=3\] mole of \[CaC{{l}_{2}}=1\]mole \[C{{a}_{3}}{{(P{{O}_{4}})}_{2}}\] \[\therefore \]0.2 mole of \[N{{a}_{3}}P{{O}_{4}}=0.3\] mole of \[CaC{{l}_{2}}\] = 0.1 mole of \[C{{a}_{3}}{{(P{{O}_{4}})}_{2}}\].
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