Question

54 g of Al and 44.8 L of O2 at NTP are reacted to form aluminium oxide. Identify the limiting reagent. Calculate the moles of product formed. Also calculate the mass of excess reactant left unused.

Answer 1

Answer:

1 mole and 16 g of O2

Explanation:

Hi!

So we have the rx as follows

4Al+3O2= 2Al2O3

This is a balanced reaction

Hence calculating no of moles in Al

{n=no of moles m= given mass W= molecular mass V=volume}

n=m/w

n=54/27

n=2 moles of Al

So calculating moles of O2

n=44.8/22.4

n=2 moles of O2

A trick for calculating LR

divide obtained moles by no of moles in the  balanced reaction

ie Al=2/4 and O2=2/3

So we know Al is LR in this rx

So now we know that 4 moles of Al give 2 moles of Al2O3

so 2 moles of Al will give 2/2 ie 1 mole

So now

we have 1 mole of Al2O3

Then ER obtained is O2

So if 4 moles of Al gives 3 moles of O2 then 2 moles of Al gives 3/4*2

ie 3/2 moles of O2

So 1.5 moles of O2 is used up then in 2 moles of O2 then 0.5 moles are left in Excess

So the mass is

0.5=given mass/ 32 g

0.5*32 ie left mass in 16 g

So unused mass of O2 is 16 g

I hope it helps !!