Question

A piece of sodium metal reacts completely with water as follows.

2 Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g)

The hydrogen gas generated is collected over water at 25.0°C. The volume of the gas is 475 mL measured at 1.00 atm. Calculate the number of grams of sodium used in the reaction. (Vapor pressure of water at 25°C = 0.0313 atm.)

Answer 1

Solution-

From the quetion we have the volume of gas V = 475 mL
= 0.475 L
and the temperature, T = 25 deg C
=298 K
the total pressure, PT = 1.0 atm
Pressure of water = 0.0313 atm(given)
Now we find the vapor pressure of gas, P = PT- vapor pressure of water
=1.0 atm - 0.0313 atm
= 0.9687 atm
Using formula n = PV / RT let's find out vapor pressure of gas

= (0.9687 atm)(0.475L) / (0.0821 Latm/mol.K)(298K)
= 0.018 mol
  

From the stoichiometric of the equation 2 moles of Na(s) are required to produce one mole of hydrogen gas
So required number of moles of Na(s) = moles of gas * 2
= 0.018 mol *2
= 0.036 mol
Now the mass of Na(s) required can be found using moles * atomic mass ofNa
= 0.036 mol *23 g/mol
= 0.828 g