Question

Question #b00ac

Answer 1

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`.