Final answer:
To determine the mass of sodium hydroxide produced when 20.0 g of sodium metal reacts with water, we need to convert the mass of sodium metal to moles and then use the mole ratio to calculate the mass of sodium hydroxide. The mass of sodium hydroxide produced is 69.993 g.
Step-by-step explanation:
In the given chemical equation: 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g), the coefficients represent the relative number of moles of each substance involved.
According to the equation, 2 moles of sodium metal (Na) react with 2 moles of water (H₂O) to produce 2 moles of sodium hydroxide (NaOH) and 1 mole of hydrogen gas (H₂).
Therefore, to determine the mass of sodium hydroxide produced, we need to convert the mass of sodium metal to moles using its molar mass (22.99 g/mol) and then convert the moles of sodium hydroxide to grams using its molar mass (39.997 g/mol).
Let's calculate:
Moles of sodium metal = mass (g) / molar mass (g/mol) = 20.0 g / 22.99 g/mol = 0.870 mol
Moles of sodium hydroxide = 2 x 0.870 mol = 1.740 mol
Mass of sodium hydroxide = moles x molar mass = 1.740 mol x 39.997 g/mol = 69.993 g