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

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:

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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

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 ? 

  1. 96

  2. 60

  3. 84

  4. 75
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

MgCO3 -> MgO + CO2. 8 g MgO = 8/40 = 0.2 mol. This requires 0.2 mol MgCO3 = 0.2 * 84 = 16.8 g. Purity = (16.8 / 20) * 100 = 84%.

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

How is percent yield calculated?

  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 $
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  1. $\cfrac{E}{T}\times 100$
  2. $\cfrac{(E-T)}{T}\times 100$
  3. $\cfrac{T}{E}\times 100$
  4. $\cfrac{(T-E)}{T}\times 100$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

$^{14} _6C\rightarrow ^{14} _7N+X$
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?

  1. 25%

  2. 50%

  3. 75%

  4. 100%

  5. Cannot be determined
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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?

  1. $33.4\ g$
  2. $40.4\ g$
  3. $43.4\ g$
  4. $3240\ g$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $ZnO = \dfrac{80}{100} \times 41 = 33.4\space g$

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

$2Ag{ NO } _{ 3 }+Cu\rightarrow Cu{ \left( { NO } _{ 3 } \right)  } _{ 2 }+2Ag$
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 $Ag=107\ g/mol$) 

  1. $74$%
  2. $47$%
  3. $89$%
  4. $65$%
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $= \dfrac{0.08}{0.108} \times 100 = 74\%$

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

$Zn+{H} _{2}{SO} _{4}\rightarrow Zn{SO} _{4}+{H} _{2}$
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?

  1. $25$%
  2. $75$%
  3. $33$%
  4. $67$%
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $= \dfrac{1.5\times 10^{-2}}{2 \times 10^{-2}} \times 100 = 75\%$

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

$4Co+3{ O } _{ 2 }\rightarrow 2C{ o } _{ 2 }{ O } _{ 3 }$
$66.8\ g$ of Cobalt reacted with oxygen and $70.50\ g$ of $C{ o } _{ 2 }{ O } _{ 3 }$ was collected after the reaction was completed. Calculate the percent yield. (At. mass of $Co=59\ g/mol$)

  1. $75$%
  2. $80$%
  3. $85$%
  4. $90$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ 4Co \space + \space 3O _2 \rightarrow 2Co _2O _3$

$4\space moles$ of cobalt produce $2\space moles$ of $Co _2O _3$

$\Rightarrow 2\space mole$  $Co \rightarrow$  $1 \space mole \space Co _2O _3$ 

$\Rightarrow 2\times 59 \rightarrow (2\times 59 + 3\times 16)$

$\Rightarrow 118\space g \space Co \rightarrow 166\space g \space Co _2O _3$

$\Rightarrow 66.8\space g \space \rightarrow (x)$

$\Rightarrow x = \dfrac{166 \times 66.8}{118} = 93.97\space g$

$\%$ Yield $= \dfrac{70.50}{93.97}\times 100 \approx 75\%$


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

A reaction produced $30.0$ grams of carbon dioxide. If the theoretical (expected) yield was $45$ grams, what is the percentage yield?

  1. $15$%
  2. $30$%
  3. $67$%
  4. $150$%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Percentage yield of a compound is the ratio of actual yield to the excepted yield of the compound.

$\%$ Yield = $ \dfrac{\text{Actual Yield}}{\text{Expected Yield}} \times 100 = \dfrac{30}{45} \times 100 = \dfrac{2}{3}\times 100 = 67\%$