How many grams of AlCl₃ are produced if 30.0 g of H₂ are produced in the following reaction: 2 Al + 6 HCl → 2 AlCl₃ + 3 H₂

Question

How many grams of AlCl₃ are produced if 30.0 g of H₂ are produced in the following reaction: 2 Al + 6 HCl → 2 AlCl₃ + 3 H₂ 

Answer 1

2 Al + 6 HCl ==> 2 AlCl₃ + 3 H₂ ... balanced equation

Use the mole ratios in the balanced equation, along with dimensional analysis to answer the question.

molar mass H₂ = 2.02 g/mol

molar mass AlCl₃ = 133 g/mol

30.0 g H₂ x 1 mol H₂ / 2.02 g x 2 mol AlCl₃ / 3 mol H₂ x 133 g AlCl₃ / mol = 1317 g AlCl₃ = 1320 g AlCl₃

Answer 2

Hello,

1) Calculate how many moles of H₂ you have (30 g/ 2.016) [2 times the MW of Hydrogen].

2) Using the mole ratio provided in the formula (for every 3 moles of H₂ produced, 2 moles of AlCl₃ will also be produced) set up a ratio of

2 mol AlCl₃/ 3 mol H₂ = x / (# from step 1)

3) Use "x" from step 2 and multiply it by the molecular weight of AlCl₃. (internet says this should be 133.34 g/mol)

Best of luck!