3g of H2 react with 29g of O2 to give H2O. .
a.calculate the maximum amount of H2O that can be formed
b. calculate the amount of the reactant which remains unreacted
Answers
3g of hydrogen will be the limiting reagent.
2H2+O2 ------->2H2O
From the above equation it is clear that 2 mole H2 react with 1 mole O2
Molar mass of H2 = 2g
Molar mass of O2= 32 g
This implies,
4 g H2 react with 32 g O2
3 g H2 reacts with = (32/4) x 3g of O2 gas
= 24 g
As the given amount of O2 is more than required therefore O2 is excess reagent and H2 is limiting reagent.
B.
2 mole of hydrogen gas reacts to form 2 mole of water molecule therefore,
4 g of H2 produces = 36 g of water
So the amount of H2O produced by 3 g H2 = (36/4) x 3
= 27 g
Hence, 27 g of water will be produced during the recation
C.
As, 24 g of oxygen has been utilised during the recation and 29 g of oxygen was supplied therefore amount of oxygen gas left is (29-24) = 5g
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Answer:
Answer:
a. 27 g water formed .
b. 5 g O₂ left .
Explanation:
Chemical reaction follows as :
2 H₂ + O₂ ⇒ 2 H₂O
2 mol 1 mol 2 mol
2 × 2 g 32 g 2 × 18 g
4 g 32 g 36 g
4 g of H₂ react with 32 g of O₂.
1 g of H₂ react with O₂ = 32 / 4 = > 8 g
3 g of H₂ react with O₂ = 8 × 3 = > 24 g.
In question O₂ given 29 g .
So , limiting reagent is H₂.
a.
1 g of H₂ react with 36 / 4 g of H₂O
3 g of H₂ form H₂O = 9 × 3 = > 27 g.
b.
Amount of the reactant i.e. O₂
= > Given amount - reactant amount
= > 29 - 24 g
= > 5 g.
Hence we get answer