Tag: percentage yield

Questions Related to percentage yield

With the amounts of reactants provided, it was possible to produce $0.667\ g$ of aspirin. One student produces $0.333\ g$ of aspirin. What was the percent yield for this student's laboratory work?

chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1. $40$%

  2. $33$%

  3. $67$%

  4. $50$%

Show answer
Correct Option: D
Explanation:

Percentage Yield of a compound is defined as ratio of actual yield to the theoretical yiald.

Actual Yield $(E) = 0.333 \space g$
Theoretical Yield $(T) = 0.667 \space g$
$\Rightarrow \% $ Yield $\dfrac{0.333}{0.667} \times 100 = 50\%$

A decomposition reaction produces sodium carbonate from sodium bicarbonate.
If the collected mass of sodium carbonate was $3.7\ g$ and the predicted amount was $4.0\ g$, what is the percent yield of the reaction?

chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1. $92.5\%$

  2. $95\%$

  3. $7.5\%$

  4. $90\%$

Show answer
Correct Option: A
Explanation:

Percent Yield of a compound is defined as ratio of actual yield to the theoretical yield.

$\Rightarrow$ Actual Yield $ = 3.7 \space g$
Theoretical Yield $ = 4.0 \space g$
So, $\% $ Yield $\dfrac{3.7}{4} \times 100$$= 92.5\%$
So, Percent Yield $= 92.5\%$

A metal oxide (MO) is reduced by heating it in a stream of hydrogen. It is found that after complete reduction, 7.95 g of oxide requires 0.2 g of $H _2$ to yield 6.35 g of the metal. We may deduce that:

chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1. The atomic weight of the metal is 48

  2. The atomic weight of the metal is 16

  3. The atomic weight of the metal is 12

  4. The atomic weight of the metal is 63.5

Show answer
Correct Option: D
Explanation:

$2MO + 2H _2 \rightarrow 2M + 2H _2O$

7.92 g    0.2g       6.35 g
let metal weight is x
mol  mol conclution 
$\dfrac{0.2}{2}$ = $\dfrac{6.35}{x}$
we get      x  =  63.5  
ans is D

20 g of a magnesium carbonate sample decomposes on heating to given carbondioxide, and 8g magnesium oxide. What will be the percentage of purity of ${\text{MgC}}{{\text{O}} _3}$ sample ? 

chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1. 96

  2. 60

  3. 84

  4. 75

Show answer
Correct Option: C

How is percent yield calculated?

chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1.  $\displaystyle   \text { Percent yield } = \dfrac { \text { actual yield } }{ \text { theoretical yield } } \times  100 $ 

  2.  $\displaystyle   \text { Percent yield } = \dfrac { \text { actual yield } }{ \text { theoretical yield }  \times  100 }  $ 

  3.  $\displaystyle   \text { Percent yield } = \dfrac { \text { actual yield } }{ \text { theoretical yield } } $

  4.  $\displaystyle   \text { Percent yield } = \dfrac {  \text { theoretical yield } }{ \text { actual yield } } $

  5.  $\displaystyle   \text { Percent yield } =  \text { actual yield } \times \text { theoretical yield  } \times  100 $

Show answer
Correct Option: A
Explanation:

$\displaystyle   \text { Percent yield } = \dfrac { \text { Actual yield } }{ \text { Theoretical yield } } \times  100 $ %

For example,
If actual yield is 10 g and theoretical yield is 20 g, 


Then, percent yield will be-$\displaystyle  \dfrac {10 g}{20g}  \times 100 = 50$ %.


Hence, option $A$ is correct.

What equation is used to calculate percent yield?
Note: $E=$ experimental; $T=$ theoretical

chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1. $\cfrac{E}{T}\times 100$

  2. $\cfrac{(E-T)}{T}\times 100$

  3. $\cfrac{T}{E}\times 100$

  4. $\cfrac{(T-E)}{T}\times 100$

Show answer
Correct Option: A
Explanation:

The calculated or expected amount of product is called theoretical yield. The amount of product actually produced is called actual yield.

When we divide actual yield by the theoretical yield and then multiplied by 100 to get the percentage yield of reaction.
$\Rightarrow \dfrac{E}{T} \times 100 = \% $ Yield

$^{14} _6C\rightarrow ^{14} _7N+X$
Water is formed by the addition of 4.0g of $H _2(g)$ to an excess of $O _2(g)$. If 27 g of $H _2O$ is recovered, what is the percent yield for the reaction?

chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1. 25%

  2. 50%

  3. 75%

  4. 100%

  5. Cannot be determined

Show answer
Correct Option: C
Explanation:

$2H _2 + O _2 \rightarrow 2H _2O$

4 g   excess   2 mol 

then water is also formed 2 mol  =  36 gram 
but it formed only 27 gram

% yeald = $\dfrac{27}{36}\times 100$ 
=  $75%$
ans is C

Consider the reaction:
$2ZnS + 3O _{2}\rightarrow 2ZnO + 2SO _{2}$
This reaction has an $80.0$ yield.
What mass of $ZnO$ is produced when $50.0\ g\ ZnS$ is heated in an open vessel untill no further weight loss is observed?

chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1. $33.4\ g$

  2. $40.4\ g$

  3. $43.4\ g$

  4. $3240\ g$

Show answer
Correct Option: A
Explanation:

Molar mass of $ZnS = 97.5\space g$

No. of moles of $ZnO = \dfrac{50}{97.5} = 0.5128\space moles$
$2\space moles$ of $ZnS$ produce $2\space moles$ of $ZnO.$
So, $0.5128\space moles$ produce $0.5128\space moles$ of $ZnO.$
So, mass of $ZnO = (0.5128)\times 81 = 41\space g$
As percentage yield $=80\%$
$\Rightarrow$ Mass of $ZnO = \dfrac{80}{100} \times 41 = 33.4\space g$

$2Ag{ NO } _{ 3 }+Cu\rightarrow Cu{ \left( { NO } _{ 3 } \right)  } _{ 2 }+2Ag$
What is the percent yield when $0.17\ g$ of $Ag{NO} _{3}$ in aqueous solution reacts with excess copper to produce $0.08\ g$ $Ag$? (At. mass of $Ag=107\ g/mol$) 

chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1. $74$%

  2. $47$%

  3. $89$%

  4. $65$%

Show answer
Correct Option: A
Explanation:

$ 2AgNO _3 \space + \space Cu \rightarrow Cu(NO _3) _2 \space + \space 2Ag $

Percentage of Ag in $AgNO _3 = \dfrac{108 \times 100}{108 + 14 + 48} = \dfrac{108}{170} \times 100 = \dfrac{1080}{17} = 63.52\%$

So, amount of Ag produced $= \dfrac{63.52}{100} \times 0.17 = 0.108\space g$

$\%$ Yield $= \dfrac{0.08}{0.108} \times 100 = 74\%$

$Zn+{H} _{2}{SO} _{4}\rightarrow Zn{SO} _{4}+{H} _{2}$
A reaction of zinc metal with sulfuric acid produces $1.5\times {10}^{-2}\ mol$ of $Zn{SO} _{4}$ from $2.0\times {10}^{-2}\ mol$ of $Zn$.
What was the percent yield of this reaction?

chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1. $25$%

  2. $75$%

  3. $33$%

  4. $67$%

Show answer
Correct Option: B
Explanation:

$1\space mole$ of Zn react with $1\space mole$ of $H _2SO _4$ produce $1\space mole$ of $ZnSO _4$.

So, to produce $1.5 \times 10^{-2} \space ZnSO _4$, $\space 1.5 \times 10^{-2} \space moles$.of zinc is needed.
Here, Actual Yield $= 1.5 \times 10^{-2} \space moles$
Theoretical Yield $= 2 \times 10^{-2} \space moles$
$\Rightarrow $ Percent Yield $= \dfrac{1.5\times 10^{-2}}{2 \times 10^{-2}} \times 100 = 75\%$