Percent Yield Answer Key (CHEM 1405)

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$\%yield=\frac{actual}{theoretical}\times100$%

Actual yield is the experimental value              Theoretical yield is the calculated value.

1. When 21.8 g of silver nitrate are reacted with an excess of sodium chloride, 17.8 g of silver chloride are formed. Calculate the percent yield of silver chloride.

AgNO₃     +     NaCl          AgCl     +     NaNO₃

Limiting Reagent                                                                                                                    Theoretical Yield
$21.8gAgNO_3\times(\frac{1molAgNO_3}{169.88gAgNO_3})\times(\frac{1molAgCl}{1molAgNO_3})\times(\frac{143.32gAgCl}{1molAgCl})=18.4gAgCl$

$\%yield=\frac{17.8gAgCl}{18.4gAgCl}\times100\%=$ 96.7%

 

 

2. Arsenic is produced by heating arsenic III oxide with carbon. Carbon dioxide is another product.   If 8.87 g of arsenic III oxide is used in the reaction and 5.33 g of arsenic is produced, what is the percent yield? 

 2As₂O₃     +     3C          4As     +     3CO₂

Limiting Reagent                                                                                                       Theoretical Yield
 $8.87gAs_2O_3\times(\frac{1molAs_2O_3}{197.84gAs_2O_3})\times(\frac{4molAs}{2molAs_2O_3})\times(\frac{74.92gAs}{1molAs})=6.72gAs$

$\%yield=\frac{5.33gAs}{6.72gAs}\times100\%=$ 79.3%

 

 

3. Diiodine pentoxide reacts with carbon monoxide to produce iodine and carbon dioxide. If 2.00 g of carbon monoxide gas is reacted, 3.17 g of iodine is produced.  Calculate the percent yield of the reaction.

I₂O₅     +     5CO          I₂     +     5CO₂

Limiting Reagent                                                                                        Theoretical Yield
 $2.00gCO\times(\frac{1molCO}{28.01gCO})\times(\frac{1molI_2}{5molCO})\times(\frac{253.8gI_2}{1molI_2})=3.62gI_2$

$\%yield=\frac{3.17gI_2}{3.62gI_2}\times100\%=$ 87.6%

 

 

 

4. Methanol (CH₃OH) can be produced through the reaction of carbon monoxide and hydrogen gas in the presence of a catalyst. If 75.0 g of carbon monoxide react to produce 68.4 g CH₃OH, what is the percent yield of CH₃OH?

CO     +     2H₂           CH₃OH

Limiting Reagent                                                                                                        Theoretical Yield
$75.0gCO\times(\frac{1molCO}{28.01gCO})\times(\frac{1molCH_3OH}{1molCO})\times(\frac{32.05gCH_3OH}{1molCH_3OH})=85.8gCH_3OH$

$\%yield=\frac{68.4gCH_3OH}{85.8gCH_3OH}\times100\%=$ 79.7%