If 2.07 grams of aluminum react with excess CuSO4, what is the theoretical yield of Cu?   2Al + 3CuSO4 ---> Al2(SO4)3 + 3Cu

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The very first thing to solve such numerical is to write a well-balanced chemical equation for the given reaction. In this case, aluminum reacts with copper sulfate, and undergoes displacement reaction, to produce aluminum sulfate and copper. The well-balanced equation is already provided by you as:

`2 Al + 3CuSO_4 -> Al_2(SO_4)_3 + 3Cu`

It is balanced, because the number of atoms of each element are same on both sides of the chemical reaction.

Now, we can use soichiometry to determine the theoretical yield of copper. From this equation, 2 moles of aluminum form 3 moles of copper.

Here, we have 2.07 g of aluminum. The molar masses of aluminum and copper are 27 g/mol and 63.5 g/mol, respectively. 

Using unitary method,

2 mole aluminum produces 3 mole copper

or, 1 mole aluminum produces 3/2 moles of copper

or, 2.07/27 moles of aluminum produces 3/2 x 2.07/27 moles of copper

= 0.115 mole copper = 0.115 x 63.5 g copper = 7.30 g copper.

Hope this helps.

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The theoretical yield of copper is the maximum amount that can be produced starting with a given quantity of a reactant, in this case 2.07 grams of aluminum.

The first step, which you've already done, is to write the balanced equation:

`2Al + 3CuSO_4 ---gt Al_2(SO_4)_3 + 3Cu`

According to the balanced equation there are 3 moles of copper produced for every two moles of aluminum used. We can calculate the theoretical yield using this ratio and the molar masses of aluminum and copper:

Al = 27.0 g/mole

Cu = 63.5 g/mole


`(2.07 g Al)/1 x (1mol)/(27.0g) x (3 mol Cu)/(2 mol Al) x (63.5 g)/(1mol) = 7.30 g Cu `

The actual yield is usually less than the theoretical yield for several reasons:

1. Reactions don't always go the completion

2. Side reactions can reduce the yield

3. It's sometimes difficult to recover all of the product

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