# How many milliliters of hydrogen will be produced by the reaction of 23.699 g Zn at 24 degrees C and 846 mm Hg?

Jul 5, 2017

You have left out a few details..........I gets approx. $8 \cdot L$ of dihydrogen gas..........

#### Explanation:

We are oxidizing zinc metal in (likely!) hydrochloric acid.....

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

And given this stoichiometry, ............................

$\text{moles of metal "-=" moles of dihydrogen}$

$\text{Moles of zinc} = \frac{23.699 \cdot g}{65.39 \cdot g \cdot m o {l}^{-} 1} = 0.362 \cdot m o l .$

And so we should generate $0.362 \cdot m o l$ $\text{dihydrogen gas}$

Now a pressure of $846 \cdot m m \cdot H g$ is specified. Clearly, the person who has set the question HAS NEVER used a mercury manometer. One atmosphere of pressure will support a column of mercury that is $760 \cdot m m$ high. And thus we must reduce the given pressure reading to units of $\text{atmospheres}$:

$\text{Pressure} = \frac{846 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} = 1.113 \cdot a t m$

And then we solve the Ideal Gas equation.......$P V = n R T$

$V = \frac{n R T}{P} = \frac{0.362 \cdot m o l \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 297.15 \cdot K}{1.113 \cdot a t m}$

$V = \text{approx." 8*L, "i.e. 8000} \cdot m L \ldots . .$

Just on the question of pressure measurement, which is why I find it objectionable that someone should quote a pressure of $846 \cdot m m$, we can use a mercury column to measure pressure at about $1 \cdot a t m$ (i.e. $760 \cdot m m \cdot H g$) or at VERY low pressures, i.e. $P = 0.01 \cdot m m \cdot H g , 5 \cdot m m \cdot H g$ etc.. You put a mercury manometer under a pressure of greater than one atmosphere, and I bet you will get mercury all over the bench, and all over the lab. This is a major clean-up job, which contract cleaners won't touch.