Question

What is the volume of 50% (w/v) H2​SO4​ required for the liberation of 5.6 lof hydrogen gas at STP on its reaction with magnesium?​

Answer 1

Here is your answer.......

Moles of H2SO4 = mass / molar mass

= 50 g / 98.076 g/mol

= 0.5098 moles H2SO4 in 100 ml solution

M = moles / L

= 0.5098 mol / 0.100 L

= 5.098 M H2SO4

1 mole of any idel gas occupies a volume of 22.4 L at STP (273 K and 1 atm)

So moles H2 you produced = 5.6 L / 22.4 L/mol

= 0.25 moles H2

balanced equation :

Mg(s) + H2SO4 ----> MgSO4(aq) + H2(g)

The balanced equation shows that 1 mole H2 forms from 1 mole H2SO4

So.... 1 : 1 ratio

Thus 0.25 moles H2 will be formed from 0.25 moles H2SO4

Find out what volume of the H2SO4 has this many moles of H2SO4 in it.

Molarity = moles / L

Therefore L = moles / molarity

= 0.25 mol / 5.098 M

= 0.049 L

= 49 ml

Hope this helps you...

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