# 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) ?

Feb 28, 2015

You'd need $\text{26.7 g}$ of zinc to produce that much hydrogen gas at those specific conditions.

So, you have your balanced chemical equation

$Z {n}_{\left(s\right)} + 2 H C {l}_{\left(a q\right)} \to Z n C {l}_{2 \left(a q\right)} + {H}_{2 \left(g\right)}$

Notice the $\text{1:1}$ 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, $P V = n R T$, to solve for the numbe of moles of hydrogen gas produced

$P V = n R T \implies n = \frac{P V}{R T} = \left(\text{1.00 atm" * "10.0 L")/(0.082("L" * "atm")/("mol" * "K") * (273.15 + 25)"K}\right)$

${n}_{{H}_{2}} = \text{0.409 moles }$ ${H}_{2}$

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

$\text{0.409 moles hydrogen" * "1 mole zinc"/"1 mole hydrogen" = "0.409 moles zinc}$

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

$\text{0.409 moles" * "65.4 g"/"1 mole" = "26.7 g zinc}$