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.

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#.