Question

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 ? (Ca = 40u, P = 31u, O = 16u)

A) 4.2 g B) 5.0 g C) 6.2 g D) 7.6 g

Answer 1

Answer:

amount of CaCl2=6.6g

moles of Na3Po4 =0.5 M

now we have to find molar mass of Na3Po4=G.M/M.M

=226/0.5

=4.52g.

=4.52/6.6

=6.2g

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