# 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 _x0001_ __ CaCO3 + __ NaNO3

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

This is a stoichiometry problem in .

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##.

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