Question

Titanium oxide `(TiO_(2))` is heated with excess hydrogen gas to give water and nee oxide `Ti_(x)O_(y)`. If `1.6 gm TiO_(2)` produces `1.44 g Ti_(x)O_(y)`, then select statement (Molar mass of titanium is `48 g//mol`):
A. Volume of `x//y` is `(2)/(3)`
B. Moles of `H_(2)` used in reaction is `0.01` mol.
C. Moles of `H_(2)` formed is `0.02` mol.
D. Both `(1)` and `(3)`

Answer 1

Correct Answer - D
`TiO_(2) + H_(2) rarr H_(2)O + Ti_(x)O_(y)`
`1.6g , 1.44g`
`:. 0.16 g` Oxygen loose from `1.6 g` oxide.
`:. 8g` Oxygen loose form `= (1.6)/(0.16) xx 8 = 80g`
Molar mass of `TiO_(2) = 48 + 32 = 80`
`TiO_(2) rarr TiO_(2-(1)/(2)) rarr TiO_(3//2) rarr Ti_(2)O_(3)`
As `0.01` mole of `O` is lost in the lost in the reaction then `H_(2)O` formed will be `0.01` mole and moles of `H_(2)` used will be also `0.01` mole.