Question

How many liters of hydrogen are obtained from the reaction of 4.00 g sodium with excess water, at STP?

Answer 1

V = 1.95 L.

Step-by-step explanation:

Hell there!

In this case, according to the following reaction between sodium metal and water:

`2Na+2H_2O\rightarrow 2NaOH+H_2`

We can realize that the moles of hydrogen can be calculated by using the initial mass of sodium, its atomic mass (23.0 g/mol) and the 2:1 mole ratio of sodium to hydrogen to obtain:

`4.00gNa*(1molNa)/(23.0gNa) *(1molH_2)/(2molNa)=0.0870molH_2`

Finally, we calculate the volume of hydrogen by using the ideal gas equation whereas the pressure is 1 atm and the temperature 273.15 K according to the STP conditions:

`PV=nRT\\\\V=(nRT)/(P)\\\\V=(0.0870mol*0.08206(atm*L)/(mol*K)*273.15 K)/(1 atm)\\\\V=1.95L`

Regards!