9.7g ofamixture of KOH and KCL was to
disolve to make 1litre solution.20cm of this solution required 25.0cm of 0.12Molarity HCL acid for complete neutralization.calculate the percentage by mass of KCL.
2 answers
I want the calculations
NOTE: You mean 20 mL or 20 cc or 20 cm^3 and not 20 cm.
KOH + HCl ==> KCl + H2O
millimols HCl = mL x M = 25 mL x 0.12 M = ?
millimols KOH = millimols HCl
mols KOH = millimols/1000 = ?
grams KOH = mols KOH x molar mass KOH = ? g KOH in the 20 mL sample. How many grams KOH in the 1000 mL? That's grams KOH in 20 cc x (1000/20) = ?
grams KOH + grams KCl = 9.7 in the sample. You know g KOH, calculate grams KCl.
% KCl = (grams KCl/grams sample)*100 = ?
Post your work if you get stuck.
KOH + HCl ==> KCl + H2O
millimols HCl = mL x M = 25 mL x 0.12 M = ?
millimols KOH = millimols HCl
mols KOH = millimols/1000 = ?
grams KOH = mols KOH x molar mass KOH = ? g KOH in the 20 mL sample. How many grams KOH in the 1000 mL? That's grams KOH in 20 cc x (1000/20) = ?
grams KOH + grams KCl = 9.7 in the sample. You know g KOH, calculate grams KCl.
% KCl = (grams KCl/grams sample)*100 = ?
Post your work if you get stuck.
1 answer