Please explain: how many grams of hydrogen are produced if...

For the reaction 2Na + 2H2O --> 2NaOH + H2, how many grams of hydrogen are produced if 52 grams of sodium and 120 grams of water are available?

PLEASE explain how to do it! I don't need the answer, but my finals are this week and I have no idea how to do this problem. I don't get why there are two given amounts in the problem. Please help!!

first you need to find the limiting reagent ...

atomic mass of sodium = 23 g/mole
molecular mass of H2O = 18 g/mole
molecular mass of H2 = 2 g/mole
molecular mass of NaOH = 40 g/mole

according to reaction ....

2 X 23 = 46 g Na reacts with 2 X 18 = 36 g H2O

so 1 g Na will react with 36/46 = 0.783 g H2O

so 52 g Na will react with 52 X 0.783 = 66.411 g H2O

so for 52 g Na we need 66.411 g H2O ...but water available is 120 g ....so clearly water is in excess and hence Na is our limiting reagent means it will entirely be consumed and amount of products will depend on amount of Na taken ....

now 2 X 23 = 46 g Na gives 2 g H2

so 1 g Na will give 2/46 = 0.043 g H2

so 52 g Na will give 52 X 0.043 = 2.236 g H2

feel free to ask any question

Right off the top of your head, you should be able to see that you have an excess of water available.

So how many moles are 52 grams of sodium? And from the balanced equation, how many moles of hydrogen are produced per mole of sodium? And how many grams is that?