12. To 100 ml CaCl2, solution containing 6.66g CaCl2, 100 ml of 0.5M Na3PO4, solution is added. The mass
of Ca3(PO4)2, precipitated would be ?
A) 4.2 g
B) 5.0 g
C) 6.2 g
D) 7.6 g
Answers
Answered by
2
Answer:
a and b and c and d
Explanation:
all are correct
Answered by
0
Answer:
Explanation:
3CaCl 2Na PO Ca (PO ) 6NaCl
Given 6.66 g 100 mL 0.5 N
=
6.66 mol
111 = 0.05 mol
= 0.06 mol
3 mol CaCl2 2 mol Na3PO4
0.06 mol CaCl2
= 3 4 3 4
2
0.06 mol Na PO 0.04 mol Na PO
3
Limiting reagent = CaCl2
3 mol CaCl2 1mol Ca3
(PO4
)
2
0.06 mol CaCl2 1/3 × 0.06 mol Ca3
(PO4
)
2
= 0.62 mol Ca3
(PO4
)
2
= 0.02 × 310 = 6.2 g
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