3.0g of a mixture of potassium carbonate and potassium chloride were dissolved in a 250cm3 standard flask. 25cm3 of this solution required 40.00cm3 of 0.01M HCl for neutralization. What is the percentage by weight of K2CO3 in the mixture? (K = 39, O = 16, C = 12)?
K2CO3 + 2 HCl --> 2 KCl + H2O + CO2
so 2 HCl mols for every K2CO3 mol
How many HCl mols did we use?
40 cm^3 * 1 liter/1000 cm^2 = .04 liters
.01 * .04 = .0004 mols of HCl
so there were 0.0002 mols of K2CO3
mol mass K2CO3 = 39*2 + 12 + 16*3 = 138 grams/mol
138 *.0002 = .0276 grams was used to neutralize
we had 250/25 or ten times that much in the 250 cm^3 flask
so .276 grams of K2CO3 were in the 250 cm^3 flask
100 * .276/3 = 9.2 percent
Thanks a lot it's really explanatory and I can always defend it if I'm asked to do so
To find the percentage by weight of K2CO3 in the mixture, we need to determine the amount of K2CO3 present in the 25 cm3 solution.
First, let's calculate the number of moles of HCl used for neutralization:
Number of moles of HCl = Concentration of HCl x Volume of HCl
= 0.01 M x 40.00 cm3
= 0.0004 moles
According to the balanced chemical equation, 1 mole of K2CO3 reacts with 2 moles of HCl. Therefore, the number of moles of K2CO3 in the 25 cm3 solution is:
Number of moles of K2CO3 = 0.0004 moles / 2
= 0.0002 moles
Next, let's calculate the molar mass of K2CO3:
Molar mass of K2CO3 = (2 x Atomic mass of K) + Atomic mass of C + (3 x Atomic mass of O)
= (2 x 39) + 12 + (3 x 16)
= 78 + 12 + 48
= 138 g/mol
Now let's calculate the mass of K2CO3 in the 25 cm3 solution:
Mass of K2CO3 = Number of moles of K2CO3 x Molar mass of K2CO3
= 0.0002 moles x 138 g/mol
= 0.0276 g
Finally, let's calculate the percentage by weight of K2CO3 in the mixture:
Percentage by weight of K2CO3 = (Mass of K2CO3 / Mass of mixture) x 100
= (0.0276 g / 3.0 g) x 100
= 0.92%
Therefore, the percentage by weight of K2CO3 in the mixture is 0.92%.
To determine the percentage by weight of K2CO3 in the mixture, we need to perform some calculations based on the given information. Here's how you can obtain the answer step by step:
Step 1: Calculate the number of moles of HCl used for neutralization.
Given that 25 cm3 of the unknown solution required 40.00 cm3 of 0.01 M HCl, we can determine the number of moles of HCl used by multiplying the volume (in liters) by the molarity (moles per liter).
Number of moles of HCl = 0.04 L (volume of HCl) × 0.01 mol/L (molarity of HCl) = 0.0004 mol.
Step 2: Establish the stoichiometric ratio between HCl and K2CO3.
From the balanced chemical equation for the reaction between HCl and K2CO3, we know that it takes 2 moles of HCl to neutralize 1 mole of K2CO3. Therefore, the number of moles of K2CO3 in the unknown solution is half the number of moles of HCl used.
Number of moles of K2CO3 = 0.0004 mol (moles of HCl) / 2 = 0.0002 mol.
Step 3: Calculate the molar mass of K2CO3.
The molar mass of K2CO3 is calculated by adding the atomic masses of its constituent elements:
K2CO3 = 39.1 g/mol (2 × atomic mass of K) + 12.01 g/mol (atomic mass of C) + 3 × 16.00 g/mol (3 × atomic mass of O) ≈ 138.21 g/mol.
Step 4: Calculate the mass of K2CO3 in the mixture.
The mass of K2CO3 in the mixture can be calculated by multiplying the number of moles of K2CO3 by its molar mass:
Mass of K2CO3 = 0.0002 mol (moles of K2CO3) × 138.21 g/mol (molar mass of K2CO3) ≈ 0.0276 g.
Step 5: Calculate the percentage by weight of K2CO3.
The percentage by weight of K2CO3 in the mixture is obtained by dividing the mass of K2CO3 by the mass of the entire mixture and multiplying by 100%:
Percentage by weight of K2CO3 = (0.0276 g / 3.0 g) × 100% ≈ 0.92%.
Therefore, the percentage by weight of K2CO3 in the mixture is approximately 0.92%.