Question

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)

Answer 1

Well, `V=(nRT)/P` and so we plug in the numbers..and approx. `12*L`...

Explanation:

We examine the reation…

`Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2`

`n_"sodium"=(23.4*g)/(22.99*g*mol^-1)=1.018*mol`...and so HALF an EQUIV dihydrogen gas is evolved ... and we have to find the appropriate gas constant...

`(1/2xx1.018*molxx8.314*L*kPa*K^-1*mol^-1xx313.4*K)/(105.3*kPa)` `-=??*L`

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

`P_"measured"=P_"dihydrogen"+P_"Saturated vapour pressure"`.

`P_"SVP"` would be extracted from a Table....and subtracted from the pressure reading to give `P_"dihydrogen"`...