Calculating concentration of product & mass of a precipitate

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To calculate the mass of calcium carbonate precipitate and the concentration of sodium nitrate produced, start with the balanced equation for the reaction between sodium carbonate and calcium nitrate. The reaction indicates that three moles of sodium carbonate react with one mole of calcium nitrate to produce three moles of calcium carbonate and two moles of sodium nitrate. Determine the moles of each reactant based on their concentrations and volumes, then identify the limiting reagent. Using the moles of the limiting reagent, calculate the mass of calcium carbonate formed and the resulting concentration of sodium nitrate in the solution. Understanding the stoichiometry of the reaction is crucial for these calculations.
lockandkey213
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The question is:
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.

I believe the balanced equation would be: 3Na2CO3 + Ca3(NO3)2 ---> 3CaCO3 + 2Na3NO3
I'm not quite sure where to go from there. Please help!
 
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There are a number of steps involved. A good idea is to write down what you know, what you don't know and how to obtain certain values (e.g. at the end, you need mass, how do you get mass from moles might be useful). Firstly, I am guessing it goes to completion? You know the number of moles that participates in a general reaction (3Na2CO3 + Ca3(NO3)2 ---> 3CaCO3 + 2Na3NO3), what does that tell you?
 
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