Question

which is cheap : 40% hydrochloric acid at rate of 50 p per kg or 80% H2SO4 at the rate of 25p per kg to cmpltly nutrliztion 7 kg of caustic potash

Answer 1

Answer:

We cna write the balanced chemical equation of the neutralization of KOH by HCl:

HCl  + KOH -> KCl + H2O

36.5g      56g

So, we get that 56 g KOH is neutralized by 36.5 g HCl .

Then the value when 7 kg KOH will be neutralized by (36.5/56) × 7 kg = 4.56 kg HCl .

So from the question the mass of 40% HCl for neutralizing 7 kg KOH will be (100/40) x 4.56  = 11.4 Kg.

Now again we neutralize the KOH by H2SO4:

H2SO4   +    2KOH     ->   K2SO4 + 2H2O

98 g              2×56 =112 g

Since, 112 g KOH is neutralized by the 98 g H2SO4.

7 kg KOH will then be nuetralized by (98/112)×7 kg = 6.12 kg of the H2SO4.

So, the form the question mass of 80% H2SO4 for neutralizing 7 Kg KOH = (100/80) × 6.12  = 7.65 kg.

Price required for 11.4 kg HCl = 11.4 × 50= Rs. 570 .

Price required for  7.65 kg H2SO4 = 7.65 × 25 = Rs. 191.2 .

Hence, from the above explanation we can conclude that 40% HCl  will be cheaper for neutralizing 7 kg  KOH.

Answer 2

Answer:

Hope this helps you...................