Percentage by weight of a mixture

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anto007
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Percentage by weight of a mixture

Post by anto007 »

Hi everyone, I have the following exercise:
"A solution containing 1.15 g of a mixture of NaOH and KOH is neutralised with 48.4 mL of 0.500 N H2SO4 solution. Calculate the percentages by weight of NaOH and KOH in the original mixture."
Would you help me to understand how to solve it? Thank you
The mind is like a parachute, it only works if it opens. -A. Einstein
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ChenBeier
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Re: Percentage by weight of a mixture

Post by ChenBeier »

two equations

m(NaOH) + m(KOH) = 1.15 g

n(NaOH) +n(KOH) = 48.4 ml * 0.5 mmol/ml

use n =m/M
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anto007
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Re: Percentage by weight of a mixture

Post by anto007 »

thanks, I've already solved it
The mind is like a parachute, it only works if it opens. -A. Einstein
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