Catherine G. answered • 03/17/19

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First, you must balance your chemical equation. We do this by making sure all the atoms on the reactant side of the equation are accounted for in the product side of the equation.

So this equation balances as 1,2 ->1,1,1

Then you need to determine which reactant is consumed first. We do this by just using that one reagent to calculate the product. Then repeat with the next reagent. Whichever produces the least amount of product will be your limiting reagent. That amount of product will be your theoretical yield.

So starting with 1000 g of CaCO3, you begin the factor/label method and multiply by 1 mole/100.09 g of CaCO3, then multiply by 1 mole CaCl2/1 mole CaCO3; and finally multiply by the molar mass of CaCl2 (which is 100.98 g/mole). This gives you an answer of 1109 g of CaCl2.

Now we look at HCl; starting with 750. g of HCl, multiply by the inverse of the molar mass (1mole/36.46g HCl); then multiply by the molar conversion (1 mole CaCl2/ 2 moles of HCl) ; and then finally by the molar mass of CaCl2 (which is 100.98 g/mole). This gives you an answer of 1141 g CaCl2.

By examining the results we see that the CaCO3 produces less product and is the limiting reagent. The theoretical yield is just 1109 g of CaCl2. There is only 4 sig fig in doing the math of that portion of the problem, so the rounding is there.

Catherine G.

Yes, but 750. Is not used in the calculation that determined the yield.03/18/19

Catherine G.

So according to the text, "General Chemistry" by Hill, et al. , "In multiplication and division, the reported result should have no more significant figures than the factor with the fewest significant figures." I have found nothing to contradict this. When looking at the multiplication line for the 1000. g answer, all the significant figures were 4 and greater, 1000. is 4, 100.09 is 5, 100.98 is 5. Therefore the accuracy of the answer is acceptable to four digits.03/18/19

J.R. S.

03/17/19