Question

A reaction takes place when 1.6 mol Al is added to 2.7 mol O2. What is present at the end of the reaction?

Answer 1

@Somy @superhelp101

Answer 2

hmm well first as we can see 1.6 is smaller - thus i'd consider Al to be limiting reagent
so using this limiting reagent mole
first find mole O2 required

Answer 3

since O2 seems to be in excess

Answer 4

how do I do that?

Answer 5

@Somy

Answer 6

ratio

Answer 7

which ratio? sorry I'm very confused haha

Answer 8

4Al + 3O2 --> 2Al2O3

Answer 9

ratio between Al and O2
look at coefficients

Answer 10

so 4al becomes 2al2 and 3O2 becomes O3 but what does that mean

Answer 11

no look
there are 2 ways of thinking of this part
1 way : work with ratio
2 way : formula

way 1

look coefficient before Al is 4 and mole of Al is 1.6
coefficient of O2 is 3 and mole - we don't know the exact amount required

thus we can put it this way

4 of Al give ---- 1.6 mole
what about 3 of O2--- X ?

so X = 3*1.6/4 = ....mole of O2 required


same thing we can do by formula
\[\frac{ mole~of~Al }{ coefficient~of~Al } = \frac{ mole~ of ~ O_2 }{ coefficient~of~O_2 }\]
thus
\[\frac{ 1.6 }{ 4 } = \frac{ mole~ of ~ O_2 }{ 3 }\]
so anyways answer is same

next after you get mole of O2 required, we can find excess mole of O2

2.7 total mole of O2 - mole that is required in reaction = Mole in excess

Answer 12

but what is present at the end of the reaction?

Answer 13

what do u mean?

Answer 14

this left over mole "that was not used" is going to be one of the products along with Al2O3

Answer 15

its not shown
but we can understand
because Al is limiting reagent (has less mole than O2 originally has)
thus the reaction depends on Al
which means Al requires specific amount of O2 to form a product Al2O3
thus there will be some O2 left by the end of the reaction which was not used

Answer 16

is it clear? if not, go ahead and ask so that i clear it out

Answer 17

It's clear. Thank you!

Answer 18

good :)
next do same but now Al with AL2O3

Answer 19

can u do it on your own now?