Please help with this question on mole concept?

9.0 gm Al, 12.0 gm sulphur and 5.6 L O_2 gas (at 0^@ C and 1 atm) are allowed to react. The maximum mass of Al_2 (SO_4)_3, which may form is -

  1. 29.0 gm
  2. 57.0 gm
  3. 42.75 gm
  4. 14.25 gm

1 Answer
Apr 10, 2018

The answer is 14.25 gm.

Explanation:

1) Write a balanced chemical reaction equation.

2Al + 3S + 6O_2rarrAl_2(SO_4)_3

2) Determine the number of moles of each reactant.

n_(Al)^0=9 gm Al(1 mol Al/26.98 gm Al) = 0.334 mol Al

n_(S)^0=12 gm S(1mol S/32.06 gm S) = 0.374 mol S

For the diatomic oxygen, we must use the ideal gas law.

n_(O_2)^o=(PV)/(RT)=(1*5.6)/(0.082057*273)=0.25 mol O_2

3) Determine the limiting reagent.

The number of starting moles of each reactant are comparable, however we require twice as many moles of diatomic oxygen than sulfur and three times as many moles of diatomic oxygen than aluminum, so this suggests that O_2 is the limiting reagent.

4) Calculate the molecular weight of the product

Molecular Weight of Al_2(SO4)_3

=2(26.98)+3(32.06)+12(16)=342.1 gm Al_2(SO4)_3 per mole

5) Calculate the number of grams of product that can be produced from the limiting reagent.

0.25 mol O_2xx (1 mol Al_2(SO_4)_3)/(6 mol O_2)xx(342.1 gm Al_2(SO_4)_3)/(mole Al_2(SO_4)_3)=14.25 gm Al_2(SO_4)_3