posted by Anonymous on .
What mass in grams of sodium hydroxide is produced if 20.0 g of sodium meal reactts with excess water according to the chemical equation 2 Na(s)+2 H2O(l)=2 NaOH(aq)+H2(g)?
I think you're supposed to use stoichiometry for this problem. This is how I would do it: 20.0g Na x (1 mol Na / molar weight Na) x (2 mols NaOH / 2 mols Na) x (molar weight NaOH / 1 mol NaOH) = ____g NaOH.
Set up an "ICE" table below the chemical equation with I being initial moles, C being change in moles, and E being end moles. Starting with I, 20.0 grams of sodium metal (20.0g x 1mol/22.990g of Na) is approximately .870 moles of sodium. We know from the equation that for every 2 moles of sodium metal reactant we get 2 moles of Sodium Hydroxide product out. So, .870 mols of Na x (2mols NaOH/2mols Na) is still .870 moles of sodium hydroxide. Now we convert back from moles to grams. .870 moles of Sodium hydroxide multiplied by it's molar mass which is 35.997grams per mole (22.990grams/mol Sodium + 15.999grams/mole Oxygen + 1.008grams/mole Hydrogen), is equal to 34.8grams of Sodium Hydroxide produced.