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

Following are the  advantages of the parametric equations: 1) they can be used to graph curves that are not functions, like the unit circle.2) the parameter can be used to represent the quantity that is  useful to  provide us with better information about the graph.3)  it is useful to convert between rectangular equations and parametric equations.with specific reference to chemistry,  parametric equations are employ for solving hydrogen and hydrogen-like ions, quantum mechanically. herein, cartesian coordinates are converted in spherical polar coordinates. the solution obtained, when plotted on spherical polar graph paper, provides visualization of atomic orbital such as s,p,d and f.