I would really appreciate some assistance with this problem. What I got thus far

Question

I would really appreciate some assistance with this problem.

What I got thus far:
2Al + 3CuSO4 --> Al2(SO4)3 + 3Cu
f 3.35 g of C are obtained from allowing 2.50 g of Cu to react with 20.00 ml of 6.0 M M what is the percen yield of the reaction? te basic equation for each calculation and show your calculation step by step to obtain full credit) (30 points) 1.1. Write balance chemical equation [Products are cu(NO3)2,NO2 and H20] 1.2. Calculate moles of Cu 1.3. Calculate moles of HNO3 1.4. Decide limiting reagent 1.5. Calculate Theoretical yield of cu(NO3)2 1.6. Calculate percent yield of cu(NO32

Explanation / Answer

1.1 ) Cu +   4HNO3 --> Cu(NO3)2 + 2NO2 + 2H2O   

is balanced eq

1.2 ) Moles of Cu = mass / atomic mass of Cu

            = 2.5 g / ( 63.546 g/mol) = 0.03934

1.3) HNO3 moles = M x V ( in L)

              = 6 x 0.02 = 0.12

1.4) as per reaction 4 HNO3 is required per 1 Cu , hence HNO3 moles required = 4 x 0.03934 =0.15736

but we had 0.12 moles of HNO3 , Hence being relatively low HNO3 is limiting reagent

1.5) Cu(NO3)2 moles expected = ( 1/4) HNO3 moles = ( 1/4) x 0.12 = 0.03

Theoretical yiled = Cu(NO3)2 moles expected x molar mass of Cu(NO3)2

              = 0.03 mol x 187.56 g/mol = 5.627 g

1.6) percent yield = ( 100 x actual / theoretical yield)

         = ( 100 x 3.35 /5.627) = 59.54 %

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