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%?
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.
To find the percentage by mass of BaCl2 in the mixture, we need to first determine the number of moles of BaCl2 and BaOH in the solution. Then we can calculate the mass of BaCl2 and use it to find the percentage.
Let's break down the problem step by step:
1. Calculate the number of moles of nitric acid used:
The volume of the nitric acid is given as 19 cm3, and its concentration is 0.1 moldm3. We can use the formula: moles = concentration × volume.
Moles of nitric acid = 0.1 moldm3 × 0.019 dm3 = 0.0019 moles.
2. Determine the reaction between BaOH and nitric acid:
From the balanced equation, we know that 1 mole of BaOH reacts with 2 moles of HNO3 to produce 1 mole of Ba(NO3)2 and 1 mole of H2O. Therefore, the number of moles of BaOH can be found by dividing the moles of nitric acid used by 2.
Moles of BaOH = 0.0019 moles ÷ 2 = 0.00095 moles.
3. Calculate the number of moles of BaCl2 reacted:
Since the molar ratio between BaOH and BaCl2 in the mixture is 1:1 (as they react in a 1:1 ratio), the number of moles of BaCl2 reacted is also 0.00095 moles.
4. Determine the molar mass of BaCl2:
The molar mass of BaCl2 can be calculated by adding the molar masses of barium (Ba) and chlorine (Cl2):
Molar mass of BaCl2 = (atomic mass of Ba) + 2 × (atomic mass of Cl) = 137.3 g/mol + 2 × 35.5 g/mol = 208.3 g/mol.
5. Calculate the mass of BaCl2:
Mass of BaCl2 = number of moles × molar mass = 0.00095 moles × 208.3 g/mol = 0.198 g.
6. Calculate the percentage by mass of BaCl2:
Percentage by mass = (mass of BaCl2 ÷ mass of mixture) × 100% = (0.198 g ÷ 2.5 g) × 100% = 7.92%.
Based on the calculations above, the correct answer is 7.92%. It seems there may have been some confusion or error in your calculations.