The HNO3 reacts only with BaIOH)2.
Ba(OH)2 + 2HNO3 ==> Ba(NO3)2 + 2H2O
mols HNO3 required = M x L = approx 0.0019
0.0019 mols HNO3 x (1 mol Ba(OH)2/2 mols HNO3) = approx 0.00095
g Ba(8OH)2 = mols x molar mass - 0.0095 x approx 171 = about 0.16 g in the 20 cc. In the 250 cc flask that is 0.16 x (250/20) = approx 2
%Ba(OH)2 = (2/2.5)*100 = about 81.2%
%BaCl2 - 100-81.2 = about 18.8. You get 18.6 if you do it more accurately than I estimated the steps.
2.5g of a mixture of anhydrous BaOH and BaCl2 was dissolved in distilled water and made up to 250cm3. 20cm3 of this solution required 19cm3 of 0.1moldm3 of dilute nitric acid for complete reaction. Calculate the percentage by mass of BaCl2 present in the mixture.
The answer is 18.6%, but I keep getting 26.6%?
1 answer