What volume of dihydrogen gas will be generated by the action of excess hydrochloric acid on a 21*g mass of zinc metal?

A temperature of $13$ ""^@C, and a pressure of $704 \cdot m m \cdot H g$ are specified...

Mar 27, 2017

To answer this question we need (i) to know that:

$\text{760 mm Hg}$ $\equiv$ $1 \cdot a t m$

Explanation:

That is $1 \cdot a t m$ of pressure will support a column of mercury that is $760 \cdot m m$ high. This is a convenient laboratory measurement that I hope has been shown to you in the lab. The difference between mercury columns can thus be related to differences between absolute pressures. A mercury column is thus useful for pressures BELOW $1 \cdot a t m$. IT IS NOT USED FOR PRESSURES ABOVE $1 \cdot a t m$. (Why not? Because you will get mercury all over the lab, and guess who is going to clean it up.)

And of course we need (ii) a stoichiometrically balanced equation:

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

So for each equiv metal, 1 equiv of dihydrogen gas results.

$\text{Moles of zinc}$ $\equiv$ $\frac{21 \cdot g}{65.38 \cdot g \cdot m o {l}^{-} 1} = 0.321 \cdot m o l$.

And thus, by the stoichiometry, $0.321 \cdot m o l$ ${H}_{2} \left(g\right)$ will result.

And so now, this is an Ideal Gas Equation, where we solve for volume:

V=(nRT)/P=(0.321*cancel(mol)xx0.0821*(L*cancel"atm")/(cancel(K^-1*mol^-1))xx286*cancelK)/((704cancel(*mm*Hg))/(760cancel(*mm*Hg*atm^-1))

You can do the math. I get an answer of approx. $8 \cdot L$ at this pressure. And this is consistent with the known molar volume of an Ideal Gas under standard conditions, i.e. approx. $25 \cdot L$.

See here for more on the use of mercury to measure moderate pressure.