FrankieV Posted June 20, 2014 Share Posted June 20, 2014 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! Link to comment Share on other sites More sharing options...
Sensei Posted June 20, 2014 Share Posted June 20, 2014 I am getting 2 grams as well. You didn't calculated sodium nitrate concentration. Link to comment Share on other sites More sharing options...
FrankieV Posted June 24, 2014 Author Share Posted June 24, 2014 Thanks That part the teacher said was unnecessary. THANKS A LOT!!! Link to comment Share on other sites More sharing options...
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