The mass of 80% pure H2SO4 required to completely neutralise 60g of NaOH is
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Answered by
112
2 Na OH + H2 SO4 => Na2 SO4 + 2 H2 O
So two moles of Na OH ie., 2 * 58.44 gms reacts with one mole ie., 98 gm of Sulfuric acid.
60 gm of hydroxide reacts with mass m of sulfuric acid
m = 60 * 98 / 116.88 gms
The mass of 80% pure acid is then: 100* m / 80
So two moles of Na OH ie., 2 * 58.44 gms reacts with one mole ie., 98 gm of Sulfuric acid.
60 gm of hydroxide reacts with mass m of sulfuric acid
m = 60 * 98 / 116.88 gms
The mass of 80% pure acid is then: 100* m / 80
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Answered by
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Answer:
The answer is 58.8 gms.
Explanation:
2NAOH+H2SO4 ->NA2SO4+2H2O
Mass of one mole of Naoh =40g.
mass of 2 moles=80g.
Mass of one mole of H2SO4=98g.
Mass of 80% pure H2SO4=80/100×98=78.4g.
Since 80 grams of Naoh reacts with 78.4g of pure H2SO4.
Mass of80% pure H2SO4 which will completely neutralise 60g of Naoh=78.4×60/80=58.8gms.
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