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"_3Ca(NO3)2+Na2CO3→CaCO3+2NaNO3
Strategy
The next problem is to convert grams of "Na"_2"CO"_3 ("A")Na2CO3(A) to litres of "Ca"("NO"_3)_2 ("B")Ca(NO3)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)_2grams of Na2CO3molar massm−−−−−−−→moles of Na2CO3molar ratiom−−−−−−−→moles of Ca(NO3)2molaritym−−−−−→litres of Ca(NO3)2
The Calculations
(a) Moles of "Na"_2"CO"_3Na2CO3
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.
