Answer:
Iron is in excess.
1) The mass of the iron remaining = 83.38 grams
2) Ethane is in excess. There will remain 90.06 grams ethane
Explanation:
Step 1: Data given
Mass of iron = 252 grams
Mass of Cl2 = 321 grams
Molar mass of Fe = 55.845
Molar mass of Cl2 = 70.9 g/mol
Step 2: The balanced equation
2Fe(s)+3Cl2(g)⟶2FeCl3(s)
Step 3: Calculate moles
Moles = mass / molar mass
Moles Fe = 252.0 grams / 55.845 g/mol = 4.512 moles
Moles Cl2 = 321.0 grams / 70.90 g/mol = 4.528 moles
Step 4: Calculate the limiting reactant
For 2 moles Fe we need 3 moles Cl2 to produce 2 moles Fecl3
Cl2 is the limiting reactant. It will completely be consumed (4.528 moles).
Fe is in excess. There will 4.528 * 2/3 = 3.019 moles be consumed
There will remain 4.512 - 3.019 = 1.493 moles of Fe
The mass of the iron remaining = 1.493 * 55.845 g/mol =83.38 grams
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If 152 grams of ethane (C2H6) are reacted with 231 grams of oxygen gas, what is the excess reactant?
Step 1: Data given
Mass of ethane = 152.0 grams
mass of O2 =231.0 grams
Molar mass of ethane = 30.07 g/mol
Molar mass of O2 = 32 g/mol
Step 2: The balanced equation
2C2H6(g) + 7O2(g) ⟶ 4CO2(g) + 6H2O(g)
Step 3: Calculate moles
Moles = mass / molar mass
Moles ethane = 152.0 grams / 30.07 g/mol = 5.055 moles
Moles O2 = 231.0 grams / 32.0 g/mol = 7.22 moles
Step 4: Calculate limiting reactant
For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O
O2 is the limiting reactant. It will completely be consumed (7.22 moles).
Ethane is in excess. There will react 7.22 * 2/7 = 2.06 moles
There will remain 5.055 - 2.06 = 2.995 moles ethane
2.995 moles ethane = 2.995 * 30.07 g/mol = 90.06 grams ethane
, we can simplify the fraction to 