Answer to Question #89963 in General Chemistry for Eileen

Question #89963
What volume of 0.250M Ca(NO3)2 is needed for a reaction that requires 12.5g of Ca(NO3)2?
1
Expert's answer
2019-05-20T08:23:18-0400

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"


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