2Al + 3CuSO4 ! Al2(SO4)3 + 3Cu
What is the maximum amount of Cu (63.5
g/mol) that could be produced by reacting
20.0 grams of Al (27.0 g/mol) with excess
CuSO4?
Here is an example stoichiometry problem I've posted. Just follow the steps. Also, remember to type the arrow somewhere so you can tell the difference between products and reactants.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Well, you see, the reaction between Al and CuSO4 is a bit like a party where Al and CuSO4 get together and exchange partners. So let's do a little "chemistry dance" to figure out how much Cu could be produced.
First, let's convert the grams of Al into moles. So, 20.0 grams of Al divided by the molar mass of Al (27.0 g/mol) gives us approximately 0.740 moles of Al.
Now we can use the balanced equation to find the mole ratio between Al and Cu. According to the equation, 2 moles of Al react with 3 moles of CuSO4 to produce 3 moles of Cu. So, since we have 0.740 moles of Al, we can expect to get approximately 1.11 moles of Cu.
Finally, we can convert moles of Cu to grams using the molar mass of Cu (63.5 g/mol). So, 1.11 moles of Cu multiplied by the molar mass of Cu gives us approximately 70.59 grams of Cu.
Therefore, the maximum amount of Cu that could be produced by reacting 20.0 grams of Al with excess CuSO4 is approximately 70.59 grams. That's quite a coppery outcome!
To determine the maximum amount of Cu that can be produced, we need to use the concept of stoichiometry. The balanced chemical equation shows that 2 moles of Al react with 3 moles of CuSO4 to produce 3 moles of Cu.
First, we need to convert the given mass of Al into moles:
n(Al) = mass(Al) / molar mass(Al)
n(Al) = 20.0 g / 27.0 g/mol
n(Al) ≈ 0.741 moles
Using the stoichiometric ratios from the balanced equation, we can calculate the moles of Cu that can be produced:
n(Cu) = (3 moles Cu / 2 moles Al) * n(Al)
n(Cu) = (3/2) * 0.741 moles
n(Cu) ≈ 1.112 moles
Finally, we can convert the moles of Cu into mass using the molar mass:
mass(Cu) = n(Cu) * molar mass(Cu)
mass(Cu) = 1.112 moles * 63.5 g/mol
mass(Cu) ≈ 70.612 g
Therefore, the maximum amount of Cu that could be produced is approximately 70.612 grams.
To find the maximum amount of Cu produced, we need to determine which reactant is limiting, meaning which reactant runs out first and limits the amount of product that can be formed.
First, we need to calculate the amount of Cu that can be produced from 20.0 grams of Al. We can use stoichiometry to do this.
The balanced equation gives us a ratio of 2 moles of Al to 3 moles of Cu. We have the molar mass of Al (27.0 g/mol), so we can convert the mass of Al to moles:
20.0 g Al x (1 mol Al / 27.0 g Al) = 0.741 mol Al
Now we can use the stoichiometric ratio to calculate the amount of Cu that can be produced:
0.741 mol Al x (3 mol Cu / 2 mol Al) = 1.112 mol Cu
To find the maximum amount of Cu in grams, we multiply the number of moles by the molar mass of Cu:
1.112 mol Cu x (63.5 g Cu / 1 mol Cu) = 70.592 g Cu
Therefore, the maximum amount of Cu that can be produced is approximately 70.592 grams.
2Al
+
3CuSO4
→
Al2(SO4)3
+
Cu