calculate mass of SO3 prodused when 500g of SO2 react with 200g of O2 according to the equation
2SO2+O2-->2SO3
identify the limiting reagent
Answers
Answer:
2SO
2
+O
2
→SO
3
Here the mole ratio between S0
2
and SO
3
is 2:2 = 1.
Here we used 2 g of SO
2
, so we need to convert that amount into moles.
Sulphur dioxide has molar mass of 64 g/mol. So,
64g/mol
2g
= 0.03125 mol of SO
2
Sulphur trioxide has molar mass of 80 g/mol. So,
0.03125 mol. 80 g/mol = 2.5 g of SO
3
Sp, 2.5 grams of sulphur trioxide was produced.
The given balanced equation is :-
2SO₂ + O₂ → 2SO₃
________________________________
Number of moles of SO₂ :-
= Given Mass/Molar mass
= 500/64
= 7.8125 moles
Number of moles of O₂ :-
= Given Mass/Molar mass
= 200/32
= 6.25 moles
________________________________
According to the balanced equation, 2 moles of SO₂ reacts with 1 mole of O₂ .
∴ 7.8125 moles of SO₂ will react with :-
= 7.8125/2
= 3.90625 moles of O₂ .
So, here we are given with more moles of O₂ than SO₂ .
Thus, SO₂ is the limiting reagent and will control the amount of product.
Now, again from the balanced equation :-
∵ 128g of SO₂ → 160g of SO₃.
∴ 500g of SO₂ → 500×160/128 = 625g of SO₃
Thus, 625g of SO₃ is produced.