Quantitative Preparation of Potassium Chloride Experimental Mass of Potassium Ch
ID: 538978 • Letter: Q
Question
Quantitative Preparation of Potassium Chloride
Experimental Mass of Potassium Chloride obtained(show calculations)
Mass of Potassium Bicarbonate after heating: 2.4604g
Moles of Potassium Bicarbonate: 0.024575 mol
Also need Experimental Moles of potassium choride obtained?
Theoretical moles of KCl?(show calculations)
Theoretical mass of KCl? (show Calculations)
Percentage error for experimental mass of KCl vs., theoretical mass of KCl? (show calculations)
Calculate the moles and grams of HCl present in the 6.0 mL of 6.0 M HCl soluton you used
If 3.000g of K2CO3 were used in this experiment (instead of KHCO3)
-What is the balanced equation for the reaction?
-How many milliliters of 6.0 M HCl would be needed?
-How many grams of KCl would be formed in this reaction?
Explanation / Answer
a)
The balanced equation fr Potassium Bicarbonate + Hydochloric acid:
KHCO3 + HCl --> KCl + H2CO3
b)
find V of HCl required
mol of KHCO3 = mass/Mw = 2.4604/100.115 = 0.02457 mol of KHCO3
so we need, 0.02457 mol of HCl as well
M = mol/V
V = mol/M = 0.02457/(6) = 0.004095 Liters = 0.004095*10^3 mL = 4.05 mL required
c)
mass of KCl formed in reaction --> ratio is 1 mol of KHCO3 per 1 mol f KCl so
0.02457 mol --> 0.02457 mol of KCl
mass = mol*MW = 0.02457 *74.5513 = 1.83172 g of KCl
Q2.
if we used K2CO3:
a)
The balanced equation fr Potassium Bicarbonate + Hydochloric acid:
K2CO3 + 2HCl --> 2KCl + H2CO3
b)
find V of HCl required
mol of K2CO3 = mass/Mw = 3/138.205 = 0.0217 mol of K2CO3
so we need, 0.0217*2 = 0.0434 mol of HCl as well
M = mol/V
V = mol/M = 0.0434/(6) = 0.007233 Liters = 0.007233*10^3 mL = 7.23 mL required
c)
mass of KCl formed in reaction --> ratio is 1 mol of K2CO3 per 2 mol f KCl so
0.02457 mol --> 0.007233 mol of KCl
mass = mol*MW = 0.007233*74.5513 = 0.539 g of KCl