Question

According to the equation 2Na + 2H2O mc012-1.jpg 2NaOH+H2, what mass of Na is required to yield 22.4 L of H2 at STP? (The atomic mass of Na is 22.99 u.). A. 1.00 g. B. 2.00 g. C. 23.0 g. D. 46.0 g

Answer 1

Option D: 46.0 g

The balanced chemical reaction is as follows:

`2Na+2H_(2)O\rightarrow 2NaOH+H_(2)`

The standard temperature and pressure conditions are 273.15 K and 1 atm respectively.

First calculate the number of moles of
`H_(2)` formed at STP,

Since,
`PV=nRT`

Here, P is pressure, V is volume, n is number of moles, R is gas constant and T is temperature.

Rearranging the equation,

`n=(PV)/(RT)=((1 atm)(22.4 L))/((0.082 atm L K^(-1) mol^(-1))(273.15 K))=1 mol`

Thus, number of moles of
`H_(2)` gas will be 1 mol.

From the chemical reaction, 2 mol of Na gives 1 mol of
`H_(2)` thus, number of moles of Na will be 2 mol.

Molar mass of Na is 22.99 g/mol thus, mass can be calculated as follows:

`m=n* M=2 mol* 22.99 g/mol=45.98 g\approx 46 g`

Therefore, mass of Na is 46.0 g.

Answer 2

we are given with the balanced equation above 2Na + 2H2O = 2NaOH + H2. when 22.4 L of H2 at STP is present, there is a one mole equivalent of H2. Via stoichiometry, there are 2 moles of Na needed. The equivalent mass of Na is equal to 45.98 grams. ANswer is D