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,

$n=\frac{PV}{RT}=\frac{(1 atm)(22.4 L)}{(0.082 atm L K^{-1} mol^{-1})(273.15 K)}=1 mol$

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:

$m=n\times M=2 mol\times 22.99 g/mol=45.98 g\approx 46 g$

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