3 genchem questions

[Dion]

I have this assignment from general chemistry which I am struggling with. The first problem reads:

-- 10 g of Al and 10 g of Br2 react according to the equation below
2Al + Br2 yields 2AlBr3
What mass of AlBr3 is formed assuming 100% yield?

...and

--- Consider the following reaction:
CH4(g) + 4Cl2(g) ® CCl4(g) + 4HCl(g)
What mass of CCl4 will be formed if 1.20 moles of methane react with 1.60 moles of chlorine?




Appreciate all the help.

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[doc toothache]

Divide the number of grams of Al by its molecular weight which will give you the number of moles of Al. (Don't forget to divide by 2 since you are dealing with 2 moles. This will give you the number of moles you expect to get for aluminum bromide. Convert the number of moles into grams.

Only 0.4 moles of methane will react with 4 moles of chlorine giving 0.4 moles of carbon tetrachloride. Divide number of moles by molecular weight to obtain the number of grams. Hope this helps.

 

[bionerd]

For the first problem:

There are 0.37 moles of Al, and 0.06 moles of Br2 available. From the reaction equation, it's clear that Al is in excess. Therefore, 0.06 moles of Br2 will react along with 0.12 moles of Al (which means there will be 0.25 moles Al left over). Since the product and Aluminum has a 1:1 ratio, this means that there will also be 0.12 moles of AlBr3 as well. Now, multiply this by its mass (266.7 g/mol) and you get 32.00 g of AlBr3.

Second problem:

Chlorine is the limiting reagent, so all of it will be used up. The ratio of CCl4:Cl2 is 1:4. So, 0.4 moles of CCl4 will be formed. Take this mole and multiply it by the mass (154.0 g/mol) and you get 61.6 g.

 

[Dion]

Thanks a million!!!