We know that, `%C = (12)/(44) xx ("Mass of " CO_(2) " formed")/("Mass of substance taken") xx 100`
Substituting the values of % of C and mass of the substance taken, we have,
`69 = (12)/(44) xx ("mass of " CO_(2) " formed")/(0.2g) xx 100` or Mass of `CO_(2)` formed `= (69 xx 44 xx 0.2)/(12 xx 100) = 0.506 g`
Similarly, `%H = (2)/(18) xx ("Mass of " H_(2)O " formed")/("Mass of substance taken") xx 100`
Substituting the values of % of H and mass of the substance taken, we have,
`4.8 = (12)/(18) xx ("Mass of " H_(2)O " formed")/(0.2) xx 100` or Mass of `H_(2)O` formed `= (4.8 xx 18 xx 0.2)/(2 xx 100) = 0.0864 g`.