Question

How may many grams of sodium hydroxide
is required to neutralise 9.8 grams sulphuric acid​

Answer 1

Explanation:

To Find -

• How many grams of NaOH is required to neutralise 9.8 g of H2SO4 ?

Solution -

Neutralisation reaction of NaOH and H2SO4 -

• 2NaOH + H2SO4 → Na2SO4 + 2H2O

Here,

molecular mass of NaOH = 23 + 16 + 1 = 40 u

molecular mass of H2SO4 = 1×2 + 32 + 16×4 = 2 + 32 + 64 = 98 u

molecular mass of Na2SO4 is 23×2 + 32 + 16×4 = 46 + 32 + 64 = 142 u

molecular mass of H2O = 1×2 + 16 = 18 u

So, we get

2NaOH + Na2SO4 → Na2SO4 + 2H2O

80g⠀⠀⠀⠀98g⠀⠀⠀⠀142g⠀⠀⠀⠀36g

Because it contains 2 mol of NaOH = 2×40 = 80 g

And it contains 2 mol of H2O = 2×18 = 36 g

Here, we see

80 g of NaOH is required to neutralise 98 g of H2SO4.

Hence,

80/98 g of NaOH is required to neutralise 1 g of H2SO4

Thus,

For 9.8 g of H2SO4 -

→ 80/98 × 9.8

→ 8 g

Therefore,

8 g of NaOH is required to neutralise 9.8 g of H2SO4.

Answer 2

Explanation:

reaction :

• 2NaOH + Na2SO4 → Na2SO4 + 2H2O

You should know:

• molecular mass of NaOH = 40 u

• molecular mass of H2SO4 =98 u

• molecular mass of Na2SO4 = 142 u

• molecular mass of H2O =18 u

solution:

2NaOH + Na2SO4 → Na2SO4 + 2H2O

80g 98g 142g 36g

80 g of NaOH is required to neutralise 98 g of H2SO4.

80/98 g of NaOH is required to neutralise 1 g of

H2SO4

⟹For 9.8 g of H2SO4

⟹80/98 x 9.8

⟹8 g

8 grams of sodium hydroxide is required to

neutralise 9.8 grams sulphuric acid