If 17.0 g Al reacts with 25.0 g O2, how much excrsss reagent will remain?
4Al + 3O2 —> 2 Al2O3
To determine how much excess reagent remains, we need to first calculate the amount of each reagent that reacts.
Step 1: Calculate the moles of Al and O2
To calculate the moles of Al, divide the given mass of Al by its molar mass:
Molar mass of Al = 26.98 g/mol
Moles of Al = Mass of Al / Molar mass of Al = 17.0 g / 26.98 g/mol
To calculate the moles of O2, divide the given mass of O2 by its molar mass:
Molar mass of O2 = 32.0 g/mol
Moles of O2 = Mass of O2 / Molar mass of O2 = 25.0 g / 32.0 g/mol
Step 2: Determine the limiting reagent
The limiting reagent is the reactant that is completely consumed in the reaction. To find the limiting reagent, we compare the mole ratio of Al to O2 from the balanced chemical equation.
From the balanced equation, the ratio of Al to O2 is 4:3.
Moles of O2 * (4 moles Al / 3 moles O2) = Moles of Al required to react with O2.
If the moles of Al required to react with O2 is less than the moles of Al given, then Al is the limiting reagent. Otherwise, O2 is the limiting reagent.
Step 3: Calculate the moles of Al2O3 formed
From the balanced equation, the ratio of Al to Al2O3 is 4:2, or simplified, 2:1.
Moles of Al * (2 moles Al2O3 / 4 moles Al) = Moles of Al2O3 formed.
Step 4: Determine the excess reagent
If Al is the limiting reagent, then we need to calculate the moles of O2 remaining.
Moles of O2 remaining = Moles of O2 given - (Moles of Al required * 3 moles O2 / 4 moles Al)
Otherwise, if O2 is the limiting reagent, then all the Al is consumed and there would be no excess reagent.
Step 5: Calculate the mass of excess reagent remaining
Mass of excess reagent remaining = Moles of excess reagent remaining * Molar mass of the excess reagent
Now you can follow these steps to calculate the amount of excess reagent that will remain in this reaction.