Chemistry, asked by gaggi2407, 10 months ago

5 gram of fuming h2 s o4 is neutralized by 26.7 ml of 0.4 n and the percent of h2 s o4 in the sample of oleum is

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Answered by Anonymous
18

Answer:

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0.5g of fuming H2SO4 is diluted with water. The solution requires 26.7ml of 0.4 N NAOH for complete neutralization. Can you find the percentage of free SO3?

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2 Answers

Renuka Varyani

Renuka Varyani, Maths Science Tutor and Executive Assistant at Sidra Medical and Research Center (2017-present)

Answered Jun 4 2018 · Author has 112 answers and 267.4k answer views

H2So4= 98/2 = 49

So3 = 80/2 = 40

26.7ml x 0.4N = 10.68ml - 0.01068L

x/49 + x/40 (0.5-X)

Eq. H2SO4 = Eq of NaOH

= (26.7 x 0.4)/1000

= 10.68 mEq.

H2SO4 + SO3 + H2O -----> 2(H2SO4)

==> SO3 + H2SO4 -----> H2SO4

∴Eq of SO3 = 1/2 of mEq. total H2SO4

==> Eq of H2SO4 + Eq of SO3 = Eq of total H2SO4

= 10.68/1000 (bec. 1 mEq. = 1/1000 Eq.)

Let x be mass of SO2 ==> mass of H2SO4

= (0.5-x)

= (0.5 - x)/49 + x/40 = 10.68/1000

Solving.... x = 0.1836g

Percentage of So3 = 0.1836/0.5 x 100

= 20.73%

Answered by ItsSpiderman44
0

Answer:

H2SO4 + SO3 + H2O -> 2(H2SO4) .

==> SO3 + H2SO4 -----> H2SO4

∴Eq of SO3 = 1/2 of H2SO4

==> Eq of H2SO4 + Eq of SO3

= Eq of total H2SO4

= 10.68/1000 (bec. 1 mEq. = 1/1000 Eq.)

Let x be mass of SO2 ==> mass of H2SO4 = (0.5-x)

= (0.5 - x)/49 + x/40 = 10.68/1000

Solving.... x = 0.1836g

Percentage = 0.1836/0.5 x 100 = 20.73%

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