3.0g of a mixture of potassium carbonate and potassium chloride were dissolve in a 250cm standard flask. 25cm of this solution required 40.00cm of 0.01mole HCl for neutralization what is the percentage by weight of K2Co3 in the mixture?
To find the percentage by weight of K2CO3 in the mixture, we need to first calculate the number of moles of K2CO3 using the volume and concentration of HCl that were required for neutralization.
1. Calculate the number of moles of HCl:
Moles of HCl = Concentration of HCl x Volume of HCl
= 0.01 mol/cm³ x 40.00 cm³
2. Determine the balanced chemical equation for the reaction between K2CO3 and HCl:
K2CO3 + 2HCl -> 2KCl + H2O + CO2
3. From the balanced equation, we can see that 1 mole of K2CO3 reacts with 2 moles of HCl. Therefore, the number of moles of K2CO3 in the solution is half the number of moles of HCl used for neutralization.
Moles of K2CO3 = 0.5 x Moles of HCl
4. Calculate the mass of K2CO3 using its molar mass:
Mass of K2CO3 = Moles of K2CO3 x Molar mass of K2CO3
= Moles of K2CO3 x (2 x Atomic mass of K + Atomic mass of C + 3 x Atomic mass of O)
5. Finally, calculate the percentage by weight of K2CO3 in the mixture:
Percentage by weight = (Mass of K2CO3 / Mass of mixture) x 100
Hope this helps! Let me know if you need any further explanation.
Note that cm means centimeter. You meant cm^3 or cubic centimeters. You can also write cc for that.
The millimols K2CO3 comes from the neutralization with HCl.
K2CO3 + 2HCl ==> 2KCl + H2O + CO2
millimols (mmol) HCl used = 40 cc HCl x 0.01 M HCl = 0.4 mmols HCl.
That 0.4 mmpl HCl was for 25 out of 250 cc original solution so that would be 4 mmols in the original 3.0 g sample. mmols K2CO3 = 1/2 mmols HCl = 2 mmols K2CO3.
mmols K2CO3 in original 3.0g = mmols K2CO3 from HCl rxn.
mmols K2CO3 in the 3.0 g sample = 1000X/molar mass K2CO3 or
(1000X/138.2) = 2 where X = mass K2CO3 in the original 3.0 g sample of the mixture of K2CO3 and KCl.
Solve for X = grams K2CO3
Then %K2CO3 = (grams K2CO3/3.0)*100 = ?
Post your work if you get stuck.