# How many liters of hydrogen gas can be produced at 313.4 K and 105.3 kPa if 23.4 g of sodium metal is completely reacted with water according to the following equation? 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

Jul 7, 2018

Well, $V = \frac{n R T}{P}$ and so we plug in the numbers..and approx. $12 \cdot L$...

#### Explanation:

We examine the reation…

$N a \left(s\right) + {H}_{2} O \left(l\right) \rightarrow N a O H \left(a q\right) + \frac{1}{2} {H}_{2}$

${n}_{\text{sodium}} = \frac{23.4 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1} = 1.018 \cdot m o l$...and so HALF an EQUIV dihydrogen gas is evolved ... and we have to find the appropriate gas constant...

$\frac{\frac{1}{2} \times 1.018 \cdot m o l \times 8.314 \cdot L \cdot k P a \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 313.4 \cdot K}{105.3 \cdot k P a}$ -=??*L

Note that given the experiment, the PRESSURE that you measure is due to....

${P}_{\text{measured"=P_"dihydrogen"+P_"Saturated vapour pressure}}$.

${P}_{\text{SVP}}$ would be extracted from a Table....and subtracted from the pressure reading to give ${P}_{\text{dihydrogen}}$...