Chem help!!?

Question

1.The reaction 2Al+3MnO---->Al2O2+3Mn proceeds until the limiting reagent is all consumed. A mixture containing 110g Al and 200g MnO was heated to initiate the reaction. Which substance is the excess reagent and by how much?

2. Calculate the theoretical yield, potassium chloride produced from the reaction of 2.56g of K and 3.85 g Cl2 if an actual experimental produces 3.81g KCL. Calculate the percentage yield.

Answer 1

2 moles of Al ( 26.98 grams /mol ) react with 3 moles of MnO ( 70.94 grams/mol )    200 grams of MnO = 200/70.94 grams/mol  =  2.82 moles .....

110 grams of Al = 110 /26.98 grams/mol = 4.08 moles

3 moles of MnO react with 2 moles of Al so 2.82 moles of MnO will react with 3/2 = 2.82/X = 1.88 moles

The Al is way in excess by 4.08-1.88 moles = 2.2 moles

2K + Cl2 ----2 KCL   2 moles of K react with 1 mole of Cl2 to give 2 moles of KCl  s****.>
2.56 grams of K = 2.56grams/39.1 grams/mol =0.0655 moles

3.85grams Cl2 = 3.85/71 grams/mole = 0.0542 moles

since K recats with Cl2 in a molar ration of 2:1   0.0655 moles of K will need only 0.0328 moles of Cl2...so the limiting reactant is K...since 2 moles of K yield 2 moles of KCl then 0.0655 moles of K will yield 0.0655 moles of KCl  soo 0.0655 moles of KCl x 74.5 grams/mole = 4.88 grams ( theoretically ) then the percentage yield will be actuall /theoretical X 100

3.81 grams/4.88 grams X 100 = 78.07%

proof

if the actual yield is 3.81 grams   that is 3.81 grams /74.5 grams /mol = 0.0511 moles ...since theroetical yield is 0.0655 moles then the percent yoeld my mole ratios is

0.0511/0.0655 X 100 = 78.02 % ( rounding errors )

Answer 2

Actually

2 moles of Al ( 26.98 grams /mol ) react with 3 moles of MnO ( 70.94 grams/mol ) 200 grams of MnO = 200/70.94 grams/mol = 2.82 moles .....

110 grams of Al = 110 /26.98 grams/mol = 4.08 moles

3 moles of MnO react with 2 moles of Al so 2.82 moles of MnO will react with 3/2 = 2.82/X = 1.88 moles

The Al is way in excess by 4.08-1.88 moles = 2.2 moles

2K + Cl2 ----2 KCL 2 moles of K react with 1 mole of Cl2 to give 2 moles of KCl s****.>
2.56 grams of K = 2.56grams/39.1 grams/mol =0.0655 moles

3.85grams Cl2 = 3.85/71 grams/mole = 0.0542 moles

since K recats with Cl2 in a molar ration of 2:1 0.0655 moles of K will need only 0.0328 moles of Cl2...so the limiting reactant is K...since 2 moles of K yield 2 moles of KCl then 0.0655 moles of K will yield 0.0655 moles of KCl soo 0.0655 moles of KCl x 74.5 grams/mole = 4.88 grams ( theoretically ) then the percentage yield will be actuall /theoretical X 100

3.81 grams/4.88 grams X 100 = 78.07%

proof

if the actual yield is 3.81 grams that is 3.81 grams /74.5 grams /mol = 0.0511 moles ...since theroetical yield is 0.0655 moles then the percent yoeld my mole ratios is

0.0511/0.0655 X 100 = 78.02 % ( rounding errors )