If 1.3 mol of aluminum reacts with excess HCl, how many moles of AlCl3 are prduced?
First write a balanced chemical equation, and then use the mole ratio as given by the coefficients2Al(s) + 6HCl(aq) - 2AlCl3(aq) + 3H2(g) 1.3 mol ?? 1.3 mol Al x (3 mol H2 / 2 mol Al) 1.95 mol H2 which you would round to 2.0 mol H2 to two significant digits Follow up Whooops!!!! Neal is correct and I went off the beam and found the moles of hydrogenThe ratio is indeed 1:1 for Al and AlCl3, and so 1.3 mols of AlCl3 would be produced, if indeed there was any AlCl3Of course we would only know that it is a 1:1 ratio for AlCl3 or a 2:3 ratio for H2 if we write the balanced chemical equationTherefore, you must still write the balanced chemical equationThe catch in problems like this is that there really is no AlCl3There are aluminum ions and chloride ions in solutionThe chloride ions are left over from the aqueous hydrochloric acidThere won't be any actual AlCl3 until all of the water is evaporated from the solutionThis is why it is helpful to write state symbols along with the formulas in balanced equationsIt's also why we like to write the net ionic equation2Al(s) + 6H+ - 2Al3+ + 3H2(g)
