If 40 liters of Chlorine react at 20 degrees Celsius and 2 atm pressure what mass of Iron III Chloride is produced?
If 60 grams of iron react completely how many molecules of iron III chloride are produced.
Finally, if 25 grams of iron react with 30 liters of Chlorine at STP what mass of iron III chloride is produced?
If 60 grams of iron react completely how many molecules of iron III chloride are produced.
Finally, if 25 grams of iron react with 30 liters of Chlorine at STP what mass of iron III chloride is produced?
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2 Fe + 3 Cl2 → 2 FeCl3
PV = nRT
n = PV / RT = (2 atm) × (40 L) / ((0.08205746 L atm/ K mol) × (20 + 273K)) = 3.327 mol Cl2
(3.327 mol Cl2) × (2/3) × (162.2048 g FeCl3/mol) = 4 × 10^2 g FeCl3
Supposing there is excess Cl2 in the reaction of 60g Fe:
(60 g Fe) / (55.8452 g Fe/mol) × (2/2) × (6.023 × 10^23 molecules/mol) =
6.5 × 10^23 molecules FeCl3
(25 g Fe) / (55.8452 g Fe/mol) = 0.4477 mol Fe
(30 L Cl2) / (22.414 L/mol) = 1.338 mol Cl2
0.4477 mole of Fe would react completely with 0.4477 × (3/2) = 0.67155 mole of Cl2, but there is more Cl2 present than that, so Cl2 is in excess and Fe is the limiting reactant.
(0.4477 mol Fe) × (2/2) × (162.2048 g FeCl3/mol) = 73 g FeCl3
PV = nRT
n = PV / RT = (2 atm) × (40 L) / ((0.08205746 L atm/ K mol) × (20 + 273K)) = 3.327 mol Cl2
(3.327 mol Cl2) × (2/3) × (162.2048 g FeCl3/mol) = 4 × 10^2 g FeCl3
Supposing there is excess Cl2 in the reaction of 60g Fe:
(60 g Fe) / (55.8452 g Fe/mol) × (2/2) × (6.023 × 10^23 molecules/mol) =
6.5 × 10^23 molecules FeCl3
(25 g Fe) / (55.8452 g Fe/mol) = 0.4477 mol Fe
(30 L Cl2) / (22.414 L/mol) = 1.338 mol Cl2
0.4477 mole of Fe would react completely with 0.4477 × (3/2) = 0.67155 mole of Cl2, but there is more Cl2 present than that, so Cl2 is in excess and Fe is the limiting reactant.
(0.4477 mol Fe) × (2/2) × (162.2048 g FeCl3/mol) = 73 g FeCl3