# The Calculation of Sodium Mass to Yield 22.4 L of H2 at STP

## Introduction

**According to the equation 2Na + 2H2O → 2NaOH + H2, what mass of Na is required to yield 22.4 L of H2 at STP?** In this calculation, we will determine the mass of sodium (Na) needed to produce 22.4 liters of hydrogen gas (H2) at Standard Temperature and Pressure (STP).

## Basic Information

At STP, the standard temperature is 273 K (0° Celsius) and the standard pressure is 1 atm. One mole of any gas at STP occupies 22.4 liters of volume (molar volume).

## Calculation

Given the equation 2Na + 2H2O → 2NaOH + H2 and the mole ratio of Na to H2 being 2:1, we can conclude that the number of moles of H2 is 1. Since one mole of H2 occupies 22.4 L at STP, the volume of the gas is already known.

Now, for the calculation of mass, we need to consider the molar mass of Na, which is 22.99 grams per mole. From the mole ratio, we can see that 2 moles of Na are needed to produce 1 mole of H2.

Therefore, the mass of Na required can be calculated by using the formula:

Mass of Na = Number of Moles of H2 x Molar Mass of Na x Mole Ratio

Mass of Na = 1 mol x 22.99 g/mol x 2 = 45.98 grams

Thus, **45.98 grams of sodium are required to yield 22.4 liters of hydrogen gas at STP.**

According to the equation 2Na + 2H2O → 2NaOH + H2, what mass of Na is required to yield 22.4 L of H2 at STP?

The mass of sodium required to yield 22.4 liters of hydrogen gas at STP is 45.98 grams.