8.4 g of a mixture of potassium hydroxide and potassium chloride were made up of 1 litre of aqueous solution. 20 cm of this solution required 24.2 cm of 0.1 m nitric acid for neutralisation. calculate the percentage by mass of potassium chloride in the mixture
8.288
To calculate the percentage by mass of potassium chloride (KCl) in the mixture, we need to determine the amount of KCl present, given the amount of nitric acid required for neutralization.
1. Convert the volume of nitric acid used for neutralization to moles:
- The molarity (M) of the nitric acid solution is 0.1 M.
- The volume used for neutralization is 24.2 cm³.
Using the formula: moles = concentration × volume
moles of nitric acid = 0.1 M × 24.2 cm³ = 2.42 mmol
2. Determine the moles of potassium hydroxide (KOH) used in the neutralization reaction:
- The stoichiometry of the reaction between nitric acid and KOH is 1:1 (1 mole of acid reacts with 1 mole of base).
- Since the moles of nitric acid used for neutralization represents the moles of KOH, we have 2.42 mmol of KOH.
3. Calculate the grams of KCl present in the mixture:
- We know the total mass of the mixture is 8.4 g.
- Subtract the mass of KOH (which is 2.42 mmol, or 56.11 grams) from the total mass of the mixture:
mass of KCl = total mass of the mixture - mass of KOH
= 8.4 g - 56.11 g
= -47.71 g (Note that the result is negative, indicating an error)
There appears to be an error in the calculations since the mass of KCl is negative. Please check the given information and re-calculate using the correct values.
KOH + HNO3 ==> KNO3 + H2O
Do you mean 1M HNO3 or 1m HNO3. There is a difference, you know. I will assume you meant 1 M. Also you must mean 20 cm^3 and not 20 cm.
mols HNO3 = M x L = 0.1M x 0.0242 L = 0.00242.
mols KOH = same = 0.00242
g KOH = mols KOH x molar mass KOH = ?
Then %KOH = (grams KOH/mass sample)*100 = ?