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Chemistry 13 Online
OpenStudy (rane):

when potassium carbonate solution is added to a solute of a calcium nitrate, a precipitate of calcium carbonate is formed. 25ml of 0.410mol/L potassium carbonate is added to 15mL of 1.05mol/L calcium nitrate solution. Calculate the mass of calcium carbonate precipitated.

OpenStudy (rane):

my q' is which moles of a substance i can use to find the mass of calcium carbonate. potassium carbonate or calcium nitrate?

OpenStudy (wolfe8):

Well, which one is the limiting reactant? Hint: It's what you have the least in proportion to what you need.

OpenStudy (rane):

potassium carbonate?

OpenStudy (wolfe8):

Idk I didn't do it myself. Lemme help. What is the balanced equation and what are the number of moles for each reactant?

OpenStudy (rane):

K2CO3 + Ca(NO3)2 ===> 2KNO3 + CaCO3

OpenStudy (wolfe8):

And the number of moles?

OpenStudy (rane):

n(K2CO3) = 0.01025mols n(Ca(NO3)2) = 0.0157mols

OpenStudy (wolfe8):

Now divide those with the number of moles required for the reaction(the constant in front of their formulas) and see which one is the smallest. That would be your limiting factor.

OpenStudy (rane):

yh ik how to find the limiting reagent. but just couldn't figure out which to use anyways thank you

OpenStudy (wolfe8):

Oh well you use the moles of the limiting agent. Limiting means it's the one that determines how much product you can produce.

OpenStudy (anonymous):

ok so the limiting reactant can be found by converting your moles and using the balanced equation as a ratio to find the precipatate which is calcium carbonate. Now it would looke like this. |dw:1386491720128:dw|

OpenStudy (anonymous):

The smallest amount is the limiting reactant.

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