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Ohh ok it's fine thank you so muchh☺️
Sorry, I meant molar mass of KOH 🤦‍♂️
Hello I was calculating the molar mass and that not the answer that I got. I get 39.99g
Ohh ok ok, thanks again

You're welcome. Let us know if you have any other questions (chem, math, etc)
Ohh ok ok, thanks again
That's the molar mass of NaOH
Thank you so much, but from where do u get the 56.1 g.
Topic updated Upwards Arrow
> 0.207

The reaction is: . . .KOH + HCl ==> H2O + NaCl

moles HCl = M HCl x L HCl = (0.118)(0.0313) = 0.00369 moles HCl

The balanced equation tells us that KOH and HCl react in a 1:1 mole ratio.

0.00369 moles HCl x (1 mole KOH / 1 mole HCl) = 0.00369 moles KOH

0.00369 moles KOH x (56.1 g NaOH / 1 mole NaOH) = 0.207 g KOH
An aliquot( 28.7 mL ) of a KOH solution required 31.3 mL of 0.118 M HCI for neutralization. What mass ( g ) of KOH was in the original sample?

A)0.129
B)0.190
C)0.173
D)0.207
E)0.215
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