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

$10g$ of limestone on heating produces $4.2g$ of $CaO$. the percentage purity of $Ca{ CO } _{ 3 }$ in limestone is: 

[Atomic mass of $Ca =$ $40$]

  1. $85%$
  2. $75%$
  3. $95%$
  4. $80%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$10g$ of limestone i.e. $CaCO _3$ contains= $\cfrac {10g}{100g/mole}$ moles of $CaCO _3=0.1$ moles of $CaCO _3$

$CaCO _3 \longrightarrow CaO+CO _2$
$1$ mole of $CaCO _3$ produce $1$ mole of $CaO$
Thus $0.1$ moles of $CaCO _3$ must produce $0.1$ mole of $CaO$
$10g$ of $CaCO _3$ must produce $0.1 \times 56= 5.6g$ of $CaO$
But $CaO$ produce is $4.2g$
Pure product obtained is $4.2g$ from $10g$ of $CaCO _3$
Product that obtain along with $1$ m purity from $10g$ of $CaCO _3$ is $5.6g$
So, percentage purity= $\cfrac {\text {mass of pure substance obtained}}{\text {mass of impure substance obtained}}\times 100$
% purity= $\cfrac {4.2}{5.6}\times 100= 75$%

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

Which of the following is not the criterion of purity of a substance?

  1. Solubility

  2. Melting point

  3. Boiling point

  4. Density
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

A pure substance shows a sharp melting point & boiling point. Hence melting point & boiling point determines the purity.

But density & solubility are not the criterion of purity of substance.

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

Consider the following reaction sequence${ CaCl } _{ 2(aq) }\quad +\quad { CO } _{ 2(g) }\quad +\quad { H } _{ 2 }O\rightarrow { CaCO } _{ 3(s) }\quad +\quad { 2HCl } _{ (aq) }$${ CaCO } _{ 3(s) }\quad \xrightarrow { heat } { CaO } _{ (s) }\quad +\quad { H } _{ 2 }{ O } _{ (g) }$if the percentage yield of the $1st$ step is $80%$ and that of the $2nd$ is $75%$, then what is the expected overall percentage yield producing $CaO$ from ${ CaCl } _{ 2 }$?

  1. $50%$
  2. $70%$
  3. $55%$
  4. $60%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
${ CaCl } _{ 2 }\longrightarrow { CaCO } _{ 3 }$            from question
$100gm\longrightarrow 80gm$
${ CaCO } _{ 3 }\longrightarrow CaO$
$100gm\longrightarrow 75gm$
$80$% $\longrightarrow 60$%
$\therefore$   The percentage of yield of $CaO$ is $60$%.
Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

Methyl benzoate is prepared by the reaction between benzoic acid and methanol, according to the equation 
$ C _{ 6 }H _{ 5 }COOH+CH _{ 3 }OH\longrightarrow C _{ 6 }H _{ 5 }COOCH _{ 3 }+H _{ 2 }O$
Benzoic acid     Methanol           Methyl benzoate
In an experiment 24.4 gm of benzoic acid were reacted with 70.0 mL of $ CH _3OH $. The density of $ CH _3OH $ is $ 0.79 g mL^{-1} $. The methyl benzoate produced had a mass of 21.6g. What was the percentage yield of product ?

  1. 91.7%

  2. 79.4%

  3. 71.5%

  4. 21.7%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Moles of benzoic acid = 24.4 / 122 = 0.20 mol. Moles of methanol = (70.0 * 0.79) / 32 = 1.728 mol. Benzoic acid is the limiting reactant. Theoretical mass of methyl benzoate = 0.20 mol * 136 g/mol = 27.2 g. Actual mass = 21.6 g. Percentage yield = (21.6 / 27.2) * 100 = 79.4 percent.

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

A sample contain $Fe\left (SO _{4}  \right ) _{3},$ $FeSO _{4a}$ and impurities. A 600 g sample contains 48g impurities ans equal moles of $Fe _{2}\left (SO _{4}.  \right )in the % of Fe _{2}\left ( SO _{4} \right ) _{3}$ in the mixture is:

  1. 33.33%

  2. 66.7%

  3. 83.33%

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The question has garbled chemical formulas (FeSO4a, SO4 instead of S, O4). Assuming it means Fe2(SO4)3 and FeSO4: Molar mass of Fe2(SO4)3 = 400 g/mol, FeSO4 = 152 g/mol. For equal moles (let n = 1), mass of Fe2(SO4)3 = 400 g, mass of FeSO4 = 152 g. Total pure = 552 g. % of Fe2(SO4)3 in pure = (400/552) × 100 = 72.5%. In total mixture (600 g): 400/600 = 66.7%. The answer depends on whether we ask % in pure mixture or total sample.

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

Percentage purity of a sample of gold is $$. How many atoms of gold are present in its $1$ gram
(Atomic mass of gold =$197 u.) 

  1. $2.6*{10^{21}}$
  2. $2.6*{10^{23}}$
  3. $3.0*{10^{21}}$
  4. $4.5*{10^{20}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Assuming 100% purity, 1 g gold = 1/197 mol. Atoms = (1/197) * 6.022 * 10^23 = 3.05 * 10^21. The provided answer 2.6 * 10^21 implies a purity of ~85%.

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

For the reaction :

     
$CaO + 2HCl \to CaC{l _2} + {H _2}O$

$2.46 g$ of CaO is reacted with excess of HCl and $3.7 g$ $ CaC{l _2}$ is formed. What is  percentage yield? 

[$Note :  \%\  Yield = \dfrac{{Actual\,yield}}{{Theoretical\,yield}} \times 100$] 

  1. 86%

  2. 26%

  3. 76%

  4. 16%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

CaO (56 g/mol): 2.46 g = 0.0439 mol. CaCl2 (111 g/mol): 3.7 g = 0.0333 mol. Theoretical yield = 0.0439 * 111 = 4.87 g. % Yield = (3.7 / 4.87) * 100 = 76%.

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

$4$ g of hydrogen $(H _2)$, $64$g of sulphur (S) and $44.8$ L of $O _2$ at STP react and form $H _2SO _4$. If $49$g of $H _2SO _4$ is formed, then $\%$ yield is  ?

  1. $25\%$
  2. $50\%$
  3. $75\%$
  4. $100\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$H _2+S+2O _2\longrightarrow H _2SO _4$

 $2g$    $32g$    $\underset {|||}{64g}$           $98g$
                  $44.8L$
The mole ratio is $H:S:O=1:1:2$
Since $O _2$ is limiting only $98g$ of $H _2SO _4$ is formed.
Given $49g$ of $H _2SO _4$ is formed.
So % yield= $\cfrac {49}{98}\times 100=50$% .

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

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?

  1. $40$%
  2. $33$%
  3. $67$%
  4. $50$%
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $\dfrac{0.333}{0.667} \times 100 = 50\%$

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

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?

  1. $92.5\%$
  2. $95\%$
  3. $7.5\%$
  4. $90\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $= 92.5\%$