Aspirin Posted April 9, 2005 Share Posted April 9, 2005 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. Link to comment Share on other sites More sharing options...
mezarashi Posted April 9, 2005 Share Posted April 9, 2005 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. Link to comment Share on other sites More sharing options...
Aspirin Posted April 9, 2005 Author Share Posted April 9, 2005 Thank you Link to comment Share on other sites More sharing options...
The Thing Posted April 9, 2005 Share Posted April 9, 2005 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. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now