Question #b00ac

1 Answer
Monzur R.
Dec 6, 2016

We need 3 "g" of "Al"_4"C"_3 to produce 1 "g" of "CH"_4

Explanation:

"Al"_4"C"_3+12"H"_2"O"rarr 4"Al"("OH")_3 + 3"CH"_4

We need one gram of methane. So we can work out how many moles we need using the moles' formula

"n"="m"/"M" = 1/16

So we need 1/16 moles of methane and using the rules of stoichiometry, we know that we need 1/48 moles of "Al"_4"C"_3. So we need to work out the mass of "Al"_4"C"_3 that gives us 1/48 moles.

"m"="nM"=1/48xx144=3 "g"

So we need 3 "g" of "Al"_4"C"_3 to produce 1 "g" of "CH"_4.

Related questions
See all questions in Introduction to Reactions and Mechanisms
Impact of this question
1445 views around the world
You can reuse this answer
Creative Commons License