Question

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Answer 1

What is the percent yield if 4.65 of copper is produced when 1.87 g of aluminum traces with an excess of copper 2 sulfate

Answer 2

The reaction here is...\[2Al^{3+}(s)+3CuSO_4(aq) \rightarrow Al_2(SO_4)_3(aq)+3Cu^{2+}(s) \]Now what we have here is 1.87g of aluminum reacting with excess copper (ii) sulfate. Let's convert that to moles...\[1.87g\ Al^{3+}*\frac{1mol}{26.982g}=0.0693mol\ Al^{3+}\]If we have that amount of aluminum, then it's going to produce a certain amount of copper when it reacts. We can figure that using a mole-ratio based on the reaction equation...\[0.0693mol\ Al*\frac{3mol\ Cu}{2mol\ Al}=0.104mol\ Cu\]Now, let's convert the actual yield of copper to an amount of copper...\[4.65g\ Cu*\frac{1mol}{63.546g}=0.0732mol\ Cu\]This is how much copper was ACTUALLY yielded, versus the amount of copper that was supposed to have theoretically been yielded. Now let's calculate the percent yield...\[\frac{0.0732}{0.104} \times 100 = 70.4\%\]