# Unlocking the Power of Stoichiometry in Chemistry

## How can we calculate the mass of Na required to yield 22.4 L of H2 at STP?

We are given with the balanced equation: 2Na + 2H2O = 2NaOH + H2. When 22.4 L of H2 at STP is present, how can we determine the mass of Na needed?

A. 1.00 g.

B. 2.00 g.

C. 23.0 g.

D. 46.0 g.

## Answer: D. 46.0 g

The balanced chemical reaction is as follows:

2Na + 2H_{2}O → 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 where P is pressure, V is volume, n is number of moles, R is gas constant, and T is temperature. Rearranging the equation, we find:

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

Thus, the 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}. Therefore, the number of moles of Na required will be 2 mol.

Since the molar mass of Na is 22.99 g/mol, we can calculate the mass as follows:

m = n x M = 2 mol x 22.99 g/mol = 45.98 g ≈ 46 g

Therefore, the mass of Na required to yield 22.4 L of H_{2} at STP is 46.0 g.