Plz explain the problem
what is the maximum number of grams of h2that can be formed from the reaction of 11.2 grams of na with 9.00 grams of h20?
the balanced equation is : 2 na + 2 > 2 naoh + h2
what is the limiting reactant?

0.250 mol H2 or 0.5 grams H2.  Convert your grams to moles (Na 11.2g = 0.5mol | 9g H2O = .5mol) then use your mole ratio (Na / H2 = 2/1 | H2O 2/1) as a proportion to calculate the # of moles produced from each reactant ( 2/1 = .5/x [solve for x]) and then convert your moles (0.25mol H2) to grams using the molar mass of H2 (H[1]x2 = 2) using multiplication. (moles x molar mass) (.250 x 2) to get your answer of 0.5 grams H2  

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