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Author Topic: chem 12  (Read 1345 times)
Sarah12345
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« on: July 02, 2009, 11:24:20 AM »

I need help on this question:

28.7 ml of a KOH solution required 31.3 mL of 0.118M HCl for neutralization.What mass (g) of KOH was in the original sample?
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kingchemist
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« Reply #1 on: July 02, 2009, 12:26:01 PM »

Write a balanced equation for the reaction and you will see that 1 mole KOH reacts with 1 mole HCl.
KOH + HCl ---> KCl + H2O
Work out moles of HCl = Volume in L x M = 0.0313 x 0.118 = ? moles
What ever this answer is, it must be the same number of moles of KOH.
So now we know the number of moles of KOH and its volume (but we don't seem to need to do anything with the volume.
So find the mass of 1 mole of KOH which is approx. 39 + 16 + 1 = 56g.
So find out what our calculated number of moles weighs. This will be the mass of KOH in 28.7 mL
« Last Edit: July 02, 2009, 12:43:23 PM by kingchemist » Logged

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