Consider the reaction
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?
1. 70.6 g
2. 57.2 g
3. 129 g
4. 31.4 g
I don't understand how to set it up nor what I do with the excess CuSO4
Well, let's break it down with a side of humor, shall we?
First, we need to find the number of moles of aluminum (Al) we have. We can use the formula:
moles = mass / molar mass
So, for 20.0 grams of Al, the number of moles can be calculated as:
moles of Al = 20.0 g / 27.0 g/mol
By crunching the numbers, we find that we have approximately 0.741 moles of Al.
According to the balanced equation, the stoichiometric ratio between Al and Cu is 2:3. So, for 0.741 moles of Al, we would need:
moles of Cu = (3/2) * moles of Al
= (3/2) * 0.741
≈ 1.111 moles
To find the maximum amount of Cu produced, we can multiply the number of moles by the molar mass of Cu:
mass of Cu = moles of Cu * molar mass of Cu
= 1.111 * 63.5 g/mol
And by doing that math, we get approximately 70.5865 grams of Cu.
So, the maximum amount of Cu that could be produced is around 70.6 grams. Well, that's a pretty cheesy amount of copper, isn't it? Option 1 sounds right to me.
To determine the maximum amount of Cu that could be produced, we need to use stoichiometry to calculate the theoretical yield of Cu. Here's how to set it up:
Step 1: Write and balance the equation for the reaction:
2 Al + 3 CuSO4 -> Al2(SO4)3 + 3 Cu
Step 2: Calculate the molar mass of Al and Cu:
Molar mass of Al = 27.0 g/mol
Molar mass of Cu = 63.5 g/mol
Step 3: Convert the mass of Al to moles:
moles of Al = mass of Al / molar mass of Al
moles of Al = 20.0 g / 27.0 g/mol ≈ 0.741 moles
Step 4: Use the stoichiometry of the reaction to find the moles of Cu produced:
From the balanced equation, we can see that 2 moles of Al react to produce 3 moles of Cu.
moles of Cu = (moles of Al) x (3 moles of Cu / 2 moles of Al)
moles of Cu = 0.741 moles x (3 moles of Cu / 2 moles of Al) = 1.1115 moles
Step 5: Convert moles of Cu to grams:
mass of Cu = moles of Cu x molar mass of Cu
mass of Cu = 1.1115 moles x 63.5 g/mol = 70.55 g
So, the maximum amount of Cu that could be produced by reacting 20.0 grams of Al with excess CuSO4 is approximately 70.55 grams.
Therefore, the correct answer is 1. 70.6 g.
To determine the maximum amount of Cu that could be produced in the reaction, you need to use stoichiometry. Stoichiometry is a method used to calculate the quantities of substances involved in a chemical reaction.
Let's set up the calculation step by step:
Step 1: Write the balanced equation for the reaction.
2Al + 3CuSO4 → Al2(SO4)3 + 3Cu
Step 2: Calculate the molar mass of Al and Cu.
Molar mass of Al = 27.0 g/mol
Molar mass of Cu = 63.5 g/mol
Step 3: Convert the given mass of Al to moles.
Given mass of Al = 20.0 grams
Number of moles of Al = Given mass / Molar mass
Number of moles of Al = 20.0 g / 27.0 g/mol ≈ 0.741 moles
Step 4: Use stoichiometry to determine the moles of Cu produced.
From the balanced equation, we can see that 2 moles of Al react with 3 moles of CuSO4 to produce 3 moles of Cu.
So, the mole ratio of Al to Cu is 2:3.
Number of moles of Cu = Number of moles of Al × (3 moles of Cu / 2 moles of Al)
Number of moles of Cu = 0.741 moles × (3/2) ≈ 1.112 moles
Step 5: Convert moles of Cu to grams.
Mass of Cu = Number of moles × Molar mass
Mass of Cu = 1.112 moles × 63.5 g/mol ≈ 70.6 grams
Therefore, the maximum amount of Cu that could be produced is approximately 70.6 grams.
So, the correct answer is option 1: 70.6 g.
The excess CuSO4 does not react.
Look at the balanced equation. For each mole of Al, 1.5 mole of CuSO4 react.
moles of AL: 20/27
Moles of CuSO4: 20/27*1.5=1.111 moles react.
Mass of reacting CuSO4: 1.111*63.5 grams.