For the reaction represented by the equation 2na + 2h2o ® 2naoh + h2, how many grams of hydrogen are produced if 120. g of sodium and 80. g of water are available?

How much of a product is made depends on the limiting reagent. there are different ways to determine the limiting reagent and this depends on the professor way of teaching it. I will use the must common one. so keep in mind that the way to solve this problem may vary. 

first we convert each grams to moles using the molar mass of the molecules. then convert moles to moles of asked molecule (H2O) using the mole-mole ration, and moles of water to grams using the molar mass of water. 

molar mass of Na= 23.0 g/mol
molar mass of H₂O= 18.0 g/mol

ratios (based on the balanced equation coefficients)
2 mol Na= 1 mol H₂
2 mol H₂O= 1 mol H₂

calculations:



since H₂O gives less of the product than the other one. H₂O is the limiting reagent and the answer is 4.49 g

Answer is to avoid experimental error