Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: jray on January 07, 2008, 09:51:57 AM
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Good day everybody.
First off all, i'm not very good at chemistry and i don't know how to solve this problem:
How many litres of 10% HCL solution (d=1.07g/cm3) it is possible to neutralise with 16 kg of NaOH?
I know it looks easy for most of you, but i couldn't solve it.
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Start from a balanced equation of neutralization reaction
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HCl(aq) + NaOH -> H2O(l) + NaCl(aq)
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Well done, now you a balanced equation. The units used in chemistry for a balanced equation is moles. You now have to convert % HCl and kg of NaOH to moles. It's just like converting imperial units to metric. (Hint: why'd they give you density of HCl? What are it's units, and how do they relate to moles?)
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M(NaOH) = 16000 x 1mol/40 = 400 moles
10g HCl solute - 100g solution
x g - 1070 g
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x=107g/l
M(HCl)=107/36=2.97mol/l
Correct me if i'm wrong
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M(NaOH) = 16000 x 1mol/40 = 400 moles
Thats OK.
10g HCl solute - 100g solution
x g - 1070 g
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x=107g/l
M(HCl)=107/36=2.97mol/l
Correct me if i'm wrong
Hard to say - you have correctly calculated amount of HCl in 1L of solution - but that's not what the questions asks for, although it can be a convenient intermediary result.
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So what i need to do next?
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Stoichiometry of reaction says you 1 mole HCl react with 1 mole of NaOH. Hence you need 400 moles of HCl.
Now you can calculate volume of HCl. (2.97 mole in 1 liter)
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Stoichiometry of reaction says you 1 mole HCl react with 1 mole of NaOH. Hence you need 400 moles of HCl.
Now you can calculate volume of HCl. (2.97 mole in 1 liter)
So just a simple division (400/2.97) or I misunderstood something?
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Just a simple division!
400 [mol] / 2,97 [mol/L] = xxx [L]
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So this is final answer? ??? Hope it is ok, thank you for the help then