Textbook M15-Calculation?

Moderators: Chem_Mod, Chem_Admin

derinceltik1K
Posts: 51
Joined: Sat Aug 17, 2019 12:15 am

Textbook M15-Calculation?

Postby derinceltik1K » Fri Oct 04, 2019 11:10 pm

Aluminum metal reacts with chlorine gas to produce aluminum chloride. In one preparation, 255 g of aluminum is placed in a container holding 535 g of chlorine gas. After reaction ceases, it is found that 300. g of aluminum chloride has been produced.

c)What is the percentage yield of aluminum chloride?

-I keep getting 67.7%. What would be the correct calculation?

Akshay Chellappa 1H
Posts: 59
Joined: Wed Sep 11, 2019 12:15 am

Re: Textbook M15-Calculation?

Postby Akshay Chellappa 1H » Fri Oct 04, 2019 11:16 pm

What did you calculate as the mass of AlCl3 produced based on the 535 grams of Cl2? For percent yield you would need to divide the actual yield, 300grams of AlCl3, by the theoretical yield, your calculation, and multiply that by 100.
Last edited by Akshay Chellappa 1H on Fri Oct 04, 2019 11:20 pm, edited 1 time in total.

Nawal Dandachi 1G
Posts: 102
Joined: Sat Sep 28, 2019 12:16 am

Re: Textbook M15-Calculation?

Postby Nawal Dandachi 1G » Fri Oct 04, 2019 11:17 pm

maybe you should recalculate because when I computed it, I got 671 grams of aluminum chloride produced. And then the percent yield would be like 44.7%

Quynh Vo
Posts: 45
Joined: Sun Sep 29, 2019 12:15 am

Re: Textbook M15-Calculation?

Postby Quynh Vo » Sat Oct 05, 2019 5:33 pm

At the end, you should’ve multiplied the 5.03 mol AlCl3 with the molar mass which is about 133.331. You have to divide the 300 g by the answer you got to get the percent yield.


Return to “Limiting Reactant Calculations”

Who is online

Users browsing this forum: No registered users and 1 guest