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1. "50cm3 of 0.1mol dm-3 solution of sulphuric acid is to be neutralised by KOH as indicated by the equation below

H2SO4 + 2KOH → K2SO4 + 2H2O

what volume of 0.05 moldm-3 KOH will be required?"
2. (Original post by tomlfc)
"50cm3 of 0.1mol dm-3 solution of sulphuric acid is to be neutralised by KOH as indicated by the equation below
H2SO4 + 2KOH → K2SO4 + 2H2O
what volume of 0.05 moldm-3 KOH will be required?"
This is not a hard question.

You know the reaction takes place in a 1:2 ratio.

Convert your sulphuric acid into mols, adjust to the ratio above, and convert back into the desired volume.
3. (Original post by Fango_Jett)
This is not a hard question.

You know the reaction takes place in a 1:2 ratio.

Convert your sulphuric acid into mols, adjust to the ratio above, and convert back into the desired volume.
how would i convert sulphuric acid into mols though?
4. (Original post by tomlfc)
how would i convert sulphuric acid into mols though?
You know the volume and concentration... n = v x c

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