Need help with some calculations

[Aspirin]

So i'm trying to make some KCL using HCl and KOH but i can not figure out how much of HCl and KOH to use.

 

Help please.

[mezarashi]

Well, this is a acid-base reaction, that will neutralize the two into water and a salt:

 

HCl(aq) + KOH(aq) -> KCl (aq) + H20 (liquid)

 

correct me if I'm wrong on the solubility

 

You will need 1 mole of HCl for every mole of KOH you have.

I'm sure you're familiar with the mole unit if you're currently doing chemistry.

You will need to make use of the periodic table and calculate the molar weight for the compound: HCl and KOH

 

HCl = 1H + 1 Cl = (1.01g/mol) + (35.45g/mol) = 36.46g/mol

KOH = 1K + 1O + 1H = (39.09g/mol) + (16g/mol) + (1.01g/mol) = 56.1g/mol

 

So the ratio of HCl and KOH would be 36.46/56.1

 

For every 36.46 grams of HCl you put into the reaction, you will need 56.1 grams of KOH to use it up.

[Aspirin]

Thank you

[The Thing]

Also don't expect 100% yield. Theoretically the yield should be 1mol of KCl = 39.09g/mol +35.45g/mol = 74.54g/mol. That is for every 36.46g of HCl and/or 56.1g of KOH you SHOULD theoretically get 74.54 g of KCl. But the percent yield will be less than 100% usually, so the actual yield should be less than 74.54g of KCl.