Question

How many grammes of zinc metal are required to produce 10.0 litres of hydrogen gas, at 25 Celsius and 1.00 atm pressure, by reaction with excess hydrochloric acid? Zn (s) + 2HCl (aq)  ZnCl2 (aq) + H2 (g) ?

Answer 1

You'd need `"26.7 g"` of zinc to produce that much hydrogen gas at those specific conditions.

So, you have your balanced chemical equation

`Zn_((s)) + 2HCl_((aq)) -> ZnCl_(2(aq)) + H_(2(g))`

Notice the `"1:1"` mole ratio you have between zinc and hydrogen gas - this means that the moles of zinc that reacted will be equal to the moles of hydrogen gas produced.

Use the ideal gas law equation, `PV = nRT`, to solve for the numbe of moles of hydrogen gas produced

`PV = nRT => n = (PV)/(RT) = ("1.00 atm" * "10.0 L")/(0.082("L" * "atm")/("mol" * "K") * (273.15 + 25)"K")`

`n_(H_2) = "0.409 moles "` `H_2`

Automatically, this also be the number of moles of zinc that reacted

`"0.409 moles hydrogen" * "1 mole zinc"/"1 mole hydrogen" = "0.409 moles zinc"`

Now just use zinc's molar mass to determine the exact mass

`"0.409 moles" * "65.4 g"/"1 mole" = "26.7 g zinc"`