If you begin with 2.7 g al and 4.05 g cl2, what mass of alcl3 can be produced?

atomic weights: Al = 26.98, Cl = 35.45 In this reaction; 2Al = 53.96 and 3Cl2 = 212.7 Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed. Step 2: (a) Ratio of Al:Cl = 2.70/4.05 = 0.6667 since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537. so Cl is limiting (b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced. From Step 1: 212.7g of Cl will produce 266.66g AlCl3 212.7g = 266.66g 4.05g = x x = 5.08g of AlCl3 can be produced (c) Al:Cl = 0.2537 Al:Cl = Al:4.05 = 0.2537 mass of Al used in reaction = 4.05 x 0.2537 = 1.027g Excess reactant = 2.70 - 1.027 = 1.67g King Leo · 9 years ago

Ok where is the image ?