Generic Stoichiometry Z (aq) + 2 Y (aq)  5 M (s) + T2 (g) Given the following information: Z = 20 g/mol Y = 10 g/mol M = 6 g/mol T = 5 g/mol If you combine 100 g of solution Z with 1.8 x 1024 molecules of solution Y: How many moles of M will precipitate out of the solution? What volume of T2 gas will be produced at STP?

Z (aq) + 2 Y (aq)  5 M (s) + T 2(g) EXCESS LIMITING Z (aq) + 2 Y (aq)  5 M (s) + T 2(g) 100 g 1.8 x 1024 molecules x mol 7.5 mol M 33.6 L T2 / 20 g /mol / 6.02 x 1023 molecules/mol 2 3 mol Y 5 x mol M = x 22.4L/mol 15 = 2x 2 : 5 3 mol Y Easy HAVE 5 mol Z x = 7.5 mol 1 2 2 : 1 1.5 mol T2 NEED 1.5 mol Z 5 mol 10 mol Y 1.5 mol SMALLER Number is LIMITING Reactant EXCESS LIMITING 1 mol Y 5 mol M x mol M = 1.8 x 1024 molecules Y = 7.5 mol M 6.02 x 1023 molecules Y 2 mol Y 1 mol T2 22.4 L T2 x L T2 = 3 mol Y = 33.6 L T2 2 mol Y 1 mol T2

Z (aq) + 2 Y (aq)  5 M (s) + T 2(g) 1.8 x 1024 molecules x mol 3 mol Y 5 mol M x mol M = 3 mol Y = 7.5 mol M 2 mol Y Y M