Can someone one help with a chemistry problem considering the following reaction: 3Fe + 4H2O → Fe3O4 + 4H2?
Roger the Mole
[Don't answer your own question. That's what the Update function in the blue "Edit" drop down menu is for.] (37.3 g Fe) / (55.8450 g Fe/mol) = 0.66792 mol Fe (42.4 g H2O) / (18.01532 g H2O/mol) = 2.35355 mol H2O 0.66792 mole of Fe would react completely with 0.66792 x (4/3) = 0.89056 mole of H2O, but there is more H2O present than that, so H2O is in excess. ((2.35355 mol H2O initially) - (0.89056 mol H2O reacted)) x (18.01532 g H2O/mol) = 26.4 g H2O left over
what is the question?
Given the reaction of 37.3 grams of Fe and 42.4 grams of water, B) Determine the grams of excess reagent that did not react. (grams of H2O?)