Answer: 0.141mole of AlCl3 will be produced
Explanation:
2Al + 3Cl2 —> 2AlCl3
Molar Mass of Cl2 = 2 x 35.5 =71g/mol
Mass conc. Of Cl2 = 15g
n = Mass conc. /molar Mass = 15/71 =0.211 mol
From the equation,
3 moles of Cl2 produced 2moles of AlCl3.
0.211mole of Cl2 will produce = (0.211x2) /3 = 0.141mole of AlCl3
Answer:
We will produce 0.141 moles of AlCl3
Explanation:
Step 1: Data given
Mass of Chlorine gas = 15.0 grams
Molar mass of Cl2 = 70.9 g/mol
Step 2: The balanced equation
2Al + 3Cl2 → 2AlCl3
Step 3: Calculate moles Cl2
Moles Cl2 = mass Cl2/ molar mass Cl2
Moles Cl2 = 15.0 grams / 70.9 g/mol
Moles Cl2 = 0.212 moles
Step 4: Calculate moles AlCl3
For 2 moles Al we need 3 moles Cl2 to produce 2 moles AlCl3
For 0.212 moles Cl2 we'll have 2/3 * 0.212 = 0.141 moles AlCl3
We will produce 0.141 moles of AlCl3
