3.3.0g 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)
How many grams? Too many decimal points.
K2CO3 + 2HCl ==> 2KCl + H2O + CO2
millimoles HCl = mL x M = 40.00 x 0.01 M = 0.4000
millimoles K2CO3 x (1 mol K2CO3/2 mols HCl) = 0.4000/2 = 0.2000 in the titrated sample. grams K2CO3 in the titrated sample = mols x molar mass = 0.0002 moles x 138.2 g/mole = 0.02764 g K2CO3 in the 25 cc sample titrated. Total amount of K2CO3 in the original sample is 0.02764 x (250 cc/25 cc) = 0.2764 grams.
Then % K2CO3 = (K2CO3 in the sample/mass sample)*100 = ?
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