Question

What volume of 0.250M Ca(NO3)2 is needed for a reaction that requires 12.5g of Ca(NO3)2?

Answer 1

Calculating number of moles of a given Ca(NO₃)₂:

$n(Ca(NO_3)_2) = \frac{m(Ca(NO_3)_2)}{M(Ca(NO_3)_2)}=\frac{12.5g}{(40 + 2\cdot 14 + 6\cdot 16)\frac{g}{mol}}=\frac{12.5g}{164\frac{g}{mol}}=0.07622mol$

From equation for molar concentration $c=\frac{n}{V}$ the volume of calcium nitrate solution can be found:

$V(Ca(NO_3)_2)=\frac{n(Ca(NO_3)_2)}{c(Ca(NO_3)_2)}=\frac{0.07622mol}{0.250\frac{mol}{L}}=0.305L=305mL$