# How do you find the mass of sodium required to produce 5.68 L of hydrogen gas STP from the reaction described by the following equation: 2Na + 2H_2O -> 2NaOH + H_2?

Mar 18, 2017

Use the balanced equation to relate the number of moles of sodium to hydrogen.

Calculate the number of moles required from the volume of hydrogen and then calculate the mass of sodium in the correct molar ratio.

#### Explanation:

The balanced equation shows that two moles of sodium will produce one mole of hydrogen gas. Using the ideal gas laws we could find that 1 mole of a gas is 22.4L volume.

Therefore, 5.68L ${H}_{2}$ is (5.68/22.4)*((L)/("L/mol")) = 0.254 "Mole" H_2

With a ratio of $\left(\frac{2}{1}\right) \left(\frac{N a}{{H}_{2}}\right)$ we will need twice as many moles of sodium, or 0.507 moles of sodium.

Multiplying this value by the molecular weight of sodium gives us the mass required:
$0.507 m o l \cdot 23 \left(\frac{g}{m o l}\right) = 11.67 g r a m s N a$