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.
Equation:
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.
To find the mass of carbon tetrachloride used, you'll need to use stoichiometry and the information given in the equation and the known yield. Let's break down the steps:
1. Start with the given mass of magnesium chloride (69.3g MgCl2).
2. Convert the mass of magnesium chloride to moles by dividing by the molar mass of MgCl2 (95.211g/mol MgCl2). This step gives you the number of moles of MgCl2.
69.3g MgCl2 * (1 mol MgCl2 / 95.211g MgCl2) = 0.7280 mol MgCl2
3. Now, use the stoichiometry of the equation to relate moles of MgCl2 to moles of carbon tetrachloride (CCl4). From the equation, you can see that 2 moles of CCl4 are produced for every 1 mole of MgCl2.
0.7280 mol MgCl2 * (2 mol CCl4 / 1 mol MgCl2) = 1.4560 mol CCl4
4. Next, convert the moles of carbon tetrachloride to grams by multiplying by the molar mass of CCl4 (153.823g CCl4/mol CCl4).
1.4560 mol CCl4 * (153.823g CCl4 / 1 mol CCl4) = 224.0573g CCl4
5. Finally, account for the known yield of 78.3% (0.783) to find the actual mass of carbon tetrachloride used. Multiply the calculated mass of CCl4 by the yield.
224.0573g CCl4 * 0.783 = 175.6165g CCl4
So, the correct answer is approximately 175.6165 grams of carbon tetrachloride were used in the process.
Without having further information about the specific issue you encountered with your calculation, it's difficult to pinpoint the exact reason why your answer was incorrect. However, based on the provided solution, you could review your calculation steps and make sure each conversion is accurately represented and the correct units are canceled out.