NAME[KEY] DATE PERIOD WS LIMITING REACTANT & PERCENT YIELD

NAME[KEY] DATE PERIOD WS LIMITING REACTANT & PERCENT YIELD




Percent (%) Yield Examples

Name:_______[KEY]_________________________ Date:___________ Period:___


WS Limiting Reactant & Percent Yield Calc’s


1. 9.00 g of Al react with an excess of H3PO4 in this reaction

2 Al(s) + 2 H3PO4(aq) → 2 AlPO4(aq) + 3 H2(g)

a) What mass of AlPO4 (121.95 g/mol) could theoretically be produced in this reaction?


9.00 g Al x 1 mol Al x 2 mol AlPO4 x 121.95 g AlPO4 = 40.7 g AlPO4

26.98 g Al 2 mol Al 1 mol AlPO4



b) What is the percent yield if you actually only recovered 30.0 g of AlPO4.


30.0 g x 100 = 73.7 %

40.7 g




2. 105 g of H3PO4 (98.00 g/mol) is reacted using the same equation from #1 above.

a) What is the theoretical yield of AlPO4 (121.95 g/mol) for this reaction?



105 g H3PO4 x 1 mol H3PO4 x 2 mol AlPO4 x 121.95 g AlPO4 = 131 g AlPO4

98.00 g H3PO4 2 mol H3PO4 1 mol AlPO4



b) What is the % yield for the reaction if you recovered 95.4 g of AlPO4?


95.4 g x 100 = 72.8 %

131 g




3. When 320 g of octane is burned in excess oxygen, 392 g of water is recovered.

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

a) Determine the theoretical yield of water.



320 g C8H18 x 1 mol C8H18 x 18 mol H2O x 18.02 g H2O = 454 g H2O

114.26 g C8H18 2 mol C8H18 1 mol H2O




b) Determine the % yield.


392 g x 100 = 86.3 %

454 g

4. 2.85 g Al(NO3)3 (213.01 g/mol) combines with 1.82 g NaOH (40.00 g/mol), and 0.996 g Al(OH)3 (78.01 g/mol) is recovered according to the following equation.

Al(NO3)3 + 3 NaOH → Al(OH)3 + 3 NaNO3

a) Determine the theoretical yield of Al(OH)3. (hint: find lim. reactant 1st and use it for yield)

2.85 g Al(NO3)3 x 1 mol Al(NO3)3 x 1 mol Al(OH)3 = 0.0134 mol Al(OH)3

213.01 g Al(NO3)3 1 mol Al(NO3)3

1.82 g NaOH x 1 mol NaOH x 1 mol Al(OH)3 = 0.0152 mol Al(OH)3

40.00 g NaOH 3 mol NaOH

0.0134 mol Al(OH)3 x 78.01 g Al(OH)3 = 1.05 g Al(OH)3

1 mol Al(OH)3


b) Determine the % yield.


0.996 g x 100 = 94.9 %

1.05 g



5. In the following reaction 3 H2 + N2 → 2 NH3

5.78 g of H2 and 6.28 g N2 are reacted, and 5.05 g of NH3 (17.04 g/mol) is recovered.


a) Determine the theoretical yield for NH3. (hint: find lim. reactant 1st and use it for yield)

5.78 g H2 x 1 mol H2 x 2 mol NH3 = 1.91 mol NH3

2.02 g H2 3 mol H2

6.28 g N2 x 1 mol N2 x 2 mol NH3 = 0.448 mol NH3

28.02 g N2 1 mol N2

0.448 mol NH3 x 17.04 g NH3 = 7.63 g NH3

1 mol NH3


b) Determine the % yield for NH3.


5.05 g x 100 = 66.2 %

7.63 g



6. A student places an iron nail (Fe) with a mass of 2.32 g into a flask containing a solution of CuSO4. The nail reacts completely, leaving a quantity of copper in the bottom of the flask. The student finds the mass of recovered copper to be 2.51 g. The equation for this reaction is

Fe + CuSO4 → FeSO4 + Cu

a) What is the percent yield?


2.32 g Fe x 1 mol Fe x 1 mol Cu x 63.55 g Cu = 2.64 g Cu

55.85 g Fe 1 mol Fe 1 mol Cu



2.51 g x 100 = 95.1 %

2.64 g

2





Tags: limiting reactant, reactant, period, percent, name[key], yield, limiting