Solutions Question 47
Question: If 0.50 mol of $ CaCl_2 $ is mixed with 0.20 mol of $ Na_3PO_4 $ , the maximum number of moles of $ Ca_3{{(PO_4)}_2} $ which can be formed, is [Pb. PMT 1998]
Options:
A) 0.70
B) 0.50
C) 0.20
D) 0.10
Show Answer
Answer:
Correct Answer: D
Solution:
$ 3,CaCl_2+2,Na_3PO_4\to Ca_3{{(PO_4)}_2}+6NaCl $
$ \therefore $ Mole of $ Na_3PO_4=3 $ mole of $ CaCl_2=1 $ mole $ Ca_3{{(PO_4)}_2} $
$ \therefore $ 0.2 mole of $ Na_3PO_4=0.3 $ mole of $ CaCl_2 $ = 0.1 mole of $ Ca_3{{(PO_4)}_2} $ .
BETA
