Question

Calculate the percent of BaO in 29.0 g of a mixture of BaO and CaO which just reacts with 100.8 mL of 6.00M HCI.
BaO + 2HCl → BaCl2 + H2O
CaO + 2HCl → CaCl2 + H2O
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Answer 1

Given info : 29.0 g of a mixture of BaO and CaO which just reacts with 100.8 mL of 6.00M HCI.

BaO + 2HCl → BaCl2 + H2O

CaO + 2HCl → CaCl2 + H2O

To find : The percentage of BaO is ...

solution : no of moles of HCl = molarity × volume in L

= 6 × 100.8/1000

= 0.6048 mol

let x g of BaO is exists in mixture then, no of moles of BaO = mass of BaO/molar mass of BaO

= x/153.33

mass of CaO = (29 - x)g

molar mass of CaO = 56g

so no of moles of CaO = (29 - x)/56

here, no of moles of BaO + no of moles of CaO = 0.6048

⇒x/153.33 + (29 - x)/56 = 0.6048/2 [ because each mole of BaO and CaO react with 2 moles of HCl ]

⇒(56x + 29 × 153.33 - 153.33x) = 0.3024 × 56 × 153.33

⇒-97.33x = 2,596.551 - 4,446.57

⇒x ≈ 19g

and 29 - x = 10 g

Therefore the percentage of BaO = mass of BaO/mass of mixture × 100

= 19/29 × 100

= 65.5 %