How many liters of a 0.75 M solution of Ca(NO3)2 will be required to react with 148 g of Na2CO3? __ Ca(NO3)2 + __ Na2CO3  __ CaCO3 + __ NaNO3

1 Answer
Sep 27, 2014

Answer:

You need 1.86 L of #"Ca"("NO"_3)_2# to react with the #"Na"_2"CO"_3#.

Explanation:

This is a stoichiometry problem in converting mass to volume.

Step 1. Write the balanced chemical equation.

The balanced chemical equation is

#"Ca"("NO"_3)_2 + "Na"_2"CO"_3 → "CaCO"_3 + "2NaNO"_3#

Strategy

The next problem is to convert grams of #"Na"_2"CO"_3 ("A")# to litres of #"Ca"("NO"_3)_2 ("B")#.

We can use the chart below to help us.

wps.prenhall.com

The process is:

#"grams of Na"_2"CO"_3 stackrelcolor(blue)("molar mass"color(white)(m)) (→) "moles of Na"_2"CO"_3 stackrelcolor(blue)("molar ratio"color(white)(m))→ "moles of Ca"("NO"_3)_2 stackrelcolor(blue)("molarity"color(white)(m))(→) "litres of Ca"("NO"_3)_2#

The Calculations

(a) Moles of #"Na"_2"CO"_3#

#148 color(red)(cancel(color(black)("g Na"_2"CO"_3))) × ("1 mol Na"_2"CO"_3)/( 105.99 color(red)(cancel(color(black)("g Na"_2"CO"_3)))) = "1.396 mol Na"_2"CO"_3 #

(b) Moles of #"Ca"("NO"_3)_3#

#1.396 color(red)(cancel(color(black)("mol Na"_2"CO"_3))) × ("1 mol Ca(NO"_3")"_2)/(1 color(red)(cancel(color(black)("mol Na"_2"CO"_3)))) = "1.396 mol Ca"("NO"_3)_3#

(c) Volume of #"Ca"("NO"_3)_2#

#1.396 color(red)(cancel(color(black)("mol Ca"("NO"_3)_3))) × ("1 L Ca"("NO"_3)_2)/(0.75 color(red)(cancel(color(black)("mol Ca"("NO"_3)_2)))) = "1.86 L Ca"("NO"_3)_2#

You need 1.86 L of #"Ca"("NO"_3)_2# to react with the #"Na"_2"CO"_3#.