Q: If 90.0 grams of sodium is dropped into 80 grams of water, how many liters of hydrogen at standard temperature and pressure would be produced?

If 90.0 grams of sodium is dropped into 80 grams of water, how many liters of hydrogen at standard temperature and pressure would be produced?

Flexi Says: The sodium reacts with water to produce sodium hydroxide and hydrogen gas. The chemical equation for the reaction is 2Na + 2H2O → 2NaOH + H2. The number of moles of Na given = 90/23 = 3.91 mol. The number of moles of water = 80/18 = 4.44 mol. Thus, Na is a limiting reagent in the reaction. Therefore, it is 3.91/2 = 1.96 mol of H2 is produced. Now 1 mol = 22.4 L, therefore 1.96 mol of H2 = 1.96 x 22.4 = 43.9 L of H2 is produced.