If 42.5 g n2 react with 10.1 g h2 according to the reaction below, how many grams of ammonia (nh3) can be produced, and how many grams of the excess reactant will be left over? unbalanced equation: n2 + h2 → nh3 be sure to show all of your work.

1) balance the equation N2 + 3 H2 → 2 NH3 

2) change the given grams to moles (42.5 g N2) / (28.01344 g N2/mol) = 1.5171 mol N2 (10.1 g H2) / (2.015894 g H2/mol) = 5.0102 mol H2 

3) compare moles * remember to multiply by the coefficient of the balanced equation 1.5171 moles of N2 would react completely with 1.5171 x (3/1) = 4.5513 moles of H2, but there is more H2 present than that, so H2 is in excess and N2 is the limiting reactant. 

4) from the balanced equation moles of limiting reactant to moles produced and then multiply by MW of product (1.5171 mol N2) x (2 mol NH3 / 1 mol N2) x (17.03056 g NH3/mol) = 51.7 g NH3 5) finally subtract the moles of limiting reactant from moles of excess reactant and multiply by excess reactants MW. ((5.0102 mol H2 initially) - (4.5513 mol H2 reacted)) x (2.015894 g H2/mol) = 0.925 g H2 left over