Here is the chemical reaction - Ca(NO3)2 + (NH4)3PO4 --> Double displacement follows... If you start with 10.25 mL of .553M Ca(NO3)2 soln, how many moles of calcium nitrate do you have in the soln? Also a part 2 in the question.

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Answer 1 · 2018-01-18T07:37:43

We got $5.67xx10^-3*mol$ with respect to calcium nitrate,,,

Bt definition... $"Concentration"="Moles of solute"/"Volume of solution"$

And thus $"moles of solute"="concentration"xx"volume"$

$=10.25xx10^-3*Lxx0.553*mol*L^-1=??*mol$

Note that this molar quantity of calcium biphosphate would precipitate from this solution.....

$Ca^(2+) + HPO_4^(2-) rarr CaHPO_4(s)darr$