Question

Can anyone please answer this question
Calculate the amount of 90%H2so4 required for the preparation of 420 kg HCl

Answer 1

HELLO FRIEND!

I am using the method of preparation with Sodium chloride.

Reaction: NaCl + H₂SO₄ ------> NaHSO₄ + HCl (Temperature should be less than 200°C).

Now, Atomic mass of
H = 1
S = 32
O = 16
Cl = 35.5

Hence, molecular mass of sulphuric acid (H₂SO₄)
= 2x1 + 32 + 4x16
= 98

Again, molecular mass of hydrochloric acid (HCl) 
= 1 + 35.5
= 36.5




Now, I will apply the unitary method.
Let the mass of H₂SO₄ (of 90% concentration) be x

So, x grams of sulphuric acid is used to prepare 420000 grams of HCl

From their molecular masses,
36.5g HCl is prepared by 98g of Sulphuric acid.

So, 420000g of HCl is prepared by =

$\frac{420000\times98}{36.5}$


$= \frac{41160000}{36.5}$

$= 1127671.23$ grams


Hence, x = 1127671.23g
or 1127.67 kg H₂SO₄ is required.


THANKS!