Question #b00ac

1 Answer
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#.