10.0 g of a mixture of BaO and CaO require 100 mL of 2.50 M HCl to react with it completely. Amount of BaO in the mixture is (At mass Ca = 40, Ba = 13

Correct Answer - A
`underset(1mol)(BaO)+underset(2mol)(2HCl)toBaCl_(2)+H_(2)O`
`underset(1mol)(CaO)+underset(2mol)(2HCl)toCaCl_(2)+H_(2)O`
Moles of HCl used `= ((100)/(1000)L)(2.5molL^(-1))`
`0.25 mol`
4 moles of HCl are used for 2 moles of mixtures.
`:.` 0.25 moles of HCl are used for 0.125 moles of mixture.
Let BaO = x mol
CaO = (0.125-x)mol
Mass of BaO `= x xx(137.34+16)g`
`= x(152.34)g`
Mass of CaO `= (0.125-x)(40+16)g`
`=(0.125-x)56g`
`153.34x+(0.125-x)56 = 10`
`97.34x = 10-7`
`x = (3)/(97.34)mol`
Mass of Bao `= (3)/(97.34)molxx153.34g mol^(-1)`
`= (3xx153.34)/(97.34)g=4.725g`
Percentage of `BaO = (4.725)/(10)xx100 = 47.25%`
`= 47.3 %`