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1) 15.2cm3 of 0.200 moledm-3 H2SO4 were neutralised by 25.0cm3 of NaOH
a} Write an equation for the reaction
b} Calculate the number of moles of H2S04 in 15.2cm3 of solution
2. (Original post by Crammyandcrabby)

1) 15.2cm3 of 0.200 moledm-3 H2SO4 were neutralised by 25.0cm3 of NaOH
a} Write an equation for the reaction
b} Calculate the number of moles of H2S04 in 15.2cm3 of solution
2NaOH + H2SO4 > Na2SO4 + 2H2O
(I think)

The two calculations for moles are:
n = m/M or n=cv, n = moles, m = mass, M = molecular mass, c = concentration, v = volume.

volume and moles have been said, so:
n = cv = 0.200 x 15.2/1000

Volume must be in dm3, so you divide cm3 by 1000.

Confer w/ someone else as well, because I would hate to give you the wrong answer.

So summary:
2NaOH + H2SO4 > Na2SO4 + 2H2O
Moles of H2SO4 = cv = 0.200 x 15.2/1000 = 0.00304 (moles)

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Updated: November 5, 2015
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