Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al (s)+3Cl2 (g)-gt;2AlCl3(s)You are given 19.0 g of aluminum and 24.0 g of chlorine gas
1Convert mass of Cl2 to moles of AlCl3 24.0g Cl2 x (1molCl2/27.0g Cl2) x (2molAlCl3/3molCl2) .592592592 .593 mol AlCl3 24.0 grams Cl2 is given in the equationI got 27.0 g Cl2 for every one mole from the Periodic chartThen the ration of 2 moles of AlCl3 for every 3 moles of Cl2 is found in the balanced equationYour grams of Cl2 and moles of Cl2 cancel out to leave you with moles of AlCl3Hope that helps!
