How much dioxygen is required to oxidize a #4*mol# quantity of aluminum metal?

2 Answers
Dwight
Dec 30, 2016

If the reaction involved is this one

#4Al(s) + 3O_2(g) rarr 2 Al_2O_3(s)#

the answer is 11.1 moles of #O_2#

Explanation:

I'm going to assume that the reaction you are referring to is

#4Al(s) + 3O_2(g) rarr 2 Al_2O_3(s)#

In which case, the ratio of #("moles of "O_2 " required")/("moles of "Al " used") = 3/4#

Write an equation that has this ratio equal to a similar one in which you use the numbers given in this problem, with #x# representing the unknown amount of #O_2#:

#("moles of "O_2 " required")/("moles of "Al " used") = 3/4 = x/14.8#

Solve for #x#

#x=(3/4) (14.8) = 11.1 " moles of "O_2#

anor277
Dec 30, 2016

We need #11.1# moles of #O_2(g)#.

Explanation:

A stoichiometric equation is required:

#2Al(s) + 3/2O_2(g) rarr Al_2O_3(s)#

Given the equation, for each equiv metal #3/4# equiv dioxygen gas are required, i.e. #14.8*molxx3/4=11.1*"mol"# #O_2# gas.

Given standard condtions, how many litres of gas does this represent?

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