100 mL of 0.2 mol/L sodium carbonate solution and 200 mL of 0.1 mol/L calcium nitrate solution are mixed together. Calculate the mass of calcium carbonate that would precipitate and the concentration of the sodium nitrate solution that will be produced.

Step 1: I wrote out the basic equation without balancing.

Na2CO3(aq)+Ca(NO3)2(aq)----->CaCO3(s)+NaNO3(aq)

Step 2: Balance.

Na2CO3(aq)+Ca(NO3)2(aq)----->CaCO3(s)+2NaNO3(aq)

Step 3: I wrote out my givens.

Compound

Na2CO3 Ca(NO3)2

Volume 0.1L 0.2L

Concentration 0.2M 0.1M

Step 4: my next step was to find limiting reagent.

C=n/V n=CV THEREFORE nNa2CO3=(0.1L)(0.2M)=0.02 moles

THEREFORE nCa(NO3)2=(0.2L)(0.1M)=0.02 moles

Is there no limiting reagent?

Step 5: I know that the mole to mole ratio of Na2CO3 to CaCO3 is 1:1 so I assumed I could just convert to grams.

n=m/mm

m=mmxn

m=(40.08+12.01+48)(0.02)

=2.0018g

=2g

IS THIS ANSWER CORRECT? IF NOT PLEASE HELP ME UNDERSTAND WHY, THANKS IN ADVANCE!

I am getting 2 grams as well.

 

You didn't calculated sodium nitrate concentration.

Thanks That part the teacher said was unnecessary. THANKS A LOT!!!