Tag: interpreting a balanced chemical reaction

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Questions Related to interpreting a balanced chemical reaction

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

To a $10$ml $1M$ aqueous solution of $Br _2$,excess of NaOH is added so that all $Br _2$ is disproportional to $Br^-$ and $BrO _3^-$, the resulting solution is freed from $Br^-$,by extraction and excess of $OH^-$ neutralised by acidifying the solution. The resulting solution is sufficient the react with $1.5$gm of impure $CaC _2I _4$ $(M=128gm /mol)$ sample. The purity by mass of Oxalate sample is the relevant reaction s are $Br _2(aq.)+OH^- \rightarrow (aq.)+BrO _3^-$
$Bro _3^-+C _2O _4^{2-}\rightarrow Br^-+CO _2$

  1. $85.3\%$
  2. $12.5\%$
  3. $90\%$
  4. $50\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Br2 + 6OH- -> 5Br- + BrO3- + 3H2O. 10 ml 1M Br2 = 0.01 mol. This produces 0.01/6 = 0.00166 mol BrO3-. Reaction: 2BrO3- + 3C2O4(2-) -> 2Br- + 6CO2. 0.00166 mol BrO3- reacts with 0.0025 mol C2O4(2-). 0.0025 mol * 128 g/mol = 0.32 g. Purity = (0.32 / 1.5) * 100 = 21.3%. The provided answer 85.3% seems to rely on different stoichiometry.

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

In the decomposition of 10 g of $Mg{ CO } _{ 3 }$, 0.1 mole ${ CO } _{ 2 }$ and 4.0 g MgO are obtained. Hence, percentage purity of $Mg{ CO } _{ 3 }$ is:

  1. 50%

  2. 60%

  3. 40%

  4. 84%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solution:- (D) $84 \%$

Molecular weight of $MgC{O} _{3} = 84 \; g$
Molecular weight of $MgO = 40 \; g$

Decomposition of $MgC{O} _{3}$-
$MgC{O} _{3} \longrightarrow MgO + C{O} _{2}$

Now, from the above reaction-
Weight of pure $MgC{O} _{3}$ required to produce $40 \; g$ of $MgO = 84.3 \; g$
Weight of pure $MgC{O} _{3}$ required to produce $4 \; g$ of $MgO = \cfrac{84.3}{40} \times 4 = 8.43 \; g$
Given weight of $MgC{O} _{3} = 10 \; g$
Now,
Amount of pure $MgC{O} _{3}$ in $10 \; g$ of given $MgC{O} _{3} = 8.43 \; g$
Thus,
Amount of pure $MgC{O} _{3}$ in $100 \; g$ of given $MgC{O} _{3} = \cfrac{8.43}{10} \times 100 = 84.3 \; g$
Therefore,
The percentage purity of given $MgC{O} _{3} = 84.3 \% \approx 84 \%$

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= $\cfrac {49}{98}\times 100=50$% .