if 90.0 grams of sodium is dropped into 80.0 g of water, how many liters of hydrogen at STP would be produced?

Answer 1

The balanced equation is:

2 Na + 2 H2O --> 2 NaOH (aq) + H2 (g)

2 moles of Sodium and 2 moles of Water gives 1 mole of H2

90.0 g / 23.0 g/mol = moles of Na

80.0 g / 18.02 g/mol = moles of water

The moles of Sodium is smaller (3.91 moles). This is the limiting reagent.

The moles of H2 produced = 3.91/2
At STP each mole has a volume of 22.4 liters, so multiply the number of moles of H2 times 22.4 L/mole to get your answer.

Answer 2

1. work out 90/23
2. work out 80/18
3. which is less?
4. use that as the limiting number of moles.
5. halve it, and multiply by 22.4

Answer 3

start up off by ability of looking out what number moles of Na are used: 2.3g/23g/mol = 0.1mol. each and each mole of Na supplies a million/2 mole of H2, subsequently: 0.1mol/2 = 0.05 mol. From this, you will locate the mass of H2 generated: 0.05mol*2g/mol = 0.1g.