Question

Alkali metals react violently with water. How much Hydrogen gas (in g) is produced when 10.0 g of sodium reacts with excess water according to the following equation?
2Na + 2H2O ----> 2NaOH + H2

Answer 1

The mass of Hydrogen gas produced when 10.0 g of sodium reacts with excess water is 0.44 g.

From the question,

We are to determine the mass of Hydrogen gas produced when 10.0g of sodium reacts with excess water.

From the balanced chemical equation for the reaction

2Na + 2H₂O → 2NaOH + H₂

This means

2 moles of sodium reacts with 2 moles of water to produce 2 moles of sodium hydroxide (NaOH) and 1 mole of Hydrogen gas.

Now, we will determine the number of moles of sodium present

Mass of sodium = 10.0 g

Atomic mass of sodium = 22.99 g/mol

From the formula

$Number\ of\ moles\ = \frac{Mass}{Molar\ mass}$

∴ Number of moles of sodium present = $\frac{10.0}{22.99}$

Number of moles of sodium present = 0.43497 mole

From the balanced chemical equation

2 moles of sodium reacts with 2 moles of water to produce 1 mole of Hydrogen gas.

Then,

0.43497 mole of sodium will react with 0.43497 mole of water to produce $\frac{0.43497}{2}$ mole of Hydrogen gas

$\frac{0.43497}{2} = 0.217485$

∴ 0.217485 mole of Hydrogen gas was produced during the reaction

Now, for the mass of Hydrogen gas produced

Using the formula

Mass = Number of moles × Molar mass

Molar mass of H₂ = 2.016 g/mol

∴ Mass of Hydrogen gas produced = 0.217485 × 2.016

Mass of Hydrogen gas produced = 0.43845 g

Mass of Hydrogen gas produced ≅ 0.44 g

Hence, the mass of Hydrogen gas produced when 10.0 g of sodium reacts with excess water is 0.44 g.

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