posted by Bach on .
If 69.3g of magnesium chloride were produced from a process with a known yield of 78.3%, find the mass of carbon tetrachloride in grams that was used.
2CCl4 + MgCrO4 -> 2COCl2 + CrO2Cl2 + MgCl2
I was able to solve it but I just don't understand why it's wrong. Here's what I put for an answer.
69.3g MgCL2 x (1 mol MgCl2/95.211g MgCl2) x (2 mol CCl4/ 1 mol MgCl2) x (153.823g CCl4/ 1 mol CCl4) x (.783/100)
Oh and after multiplying it all I got 175 g CCl4 as an answer.
I see two things wrong, one which is just a typo, I think, and the other which is why the problem is wrong.
1. The typo is you wrote .783/100 when you meant 78.3/100. You value of 175 shows you didn't use the 100 and 175 is right (for the numbers you have).
2. The problem is this.
You should calculate everything EXCEPT the percent, and I get something like 223 or 224 (you can do it more accurately) for that part of the problem. Then (divide, not multiply)
223/0.783 = about 286 grams.
I think it makes more sense to approach slightly different.
percent yield = actual yield/theoretical yield.
theo yield = actual yield/0.783
For an actual yield of 69.3 g MgCl2, it really was about 88 g if it had been at 100%. Then you go through the stoichiometry and the end of the problem tells you how much CCl4 was required to produce the 88 g MgCl2 if everything was at 100%. I get something like 285 or so.