Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:2Al + 3Cl_2 yields 2AlCl_3You are given 13.0 of aluminum and 18.0 of chlorine gas.
knowing that 1 mole of AlCl3 has within it 1 Mole of Al using molar masses 13.0 g Al (133.34 g/mol AlCl3) / ( 26.98 g/mol Al) 64.25 g AlCl3 your answer, rounded to 3 sig figs is 64.2 g AlCl3 knowing that 1 mole of AlCl3 has within it 3 Moles of Cl using molar masses 18.0 g Chlorine (133.34 g/mol AlCl3) / (3 mol Cl) (35.45 g/mol Cl) 48.90 g AlCl3 your answer, rounded to 2 sig figs is 48.9 g AlCl3 so chlorine is your limiting reagent Al was your excess reagentgiven 13.0 of aluminum and 18.0 of chlorine gaswill only produce 48.9 g AlCl3
