In the reaction `Zn(s) + 2H^(+)(aq.) rarr Zn^(2+)(aq.) + H_(2) (g)`, how many liters of hydrogen gas measured at `STP` is produced when `6.54g` of `Zn` is used `(Zn = 65.4 u)` ?

Options

A. `22.4 L`
B. `11.2 L`
C. `2.24 L`
D. `1.12 L`

Answer

C

Explanation

According to the equation, 1 mol `Zn` releases 1 mol `H_(2)`. Thus,
`n_(H_(2)) = n_(Zn) = (6.54g)/(65.4g) = 0.1 mol`
Which occupies `2.24 L` at `STP`.