Tag: stoichiometric calculations

Questions Related to stoichiometric calculations

$0.5\ g$ of impure ammonium chloride was heated with caustic soda solution to evolve ammonia gas, the gas is absorbed in $150\ mL$ of $N/5\ H _{2}SO _{4}$ solution. Excess sulphuric acid required $20\ mL$ of $1\ N\ NaOH$ for complete neutralization. The percentage of $NH _{3}$ in the ammonium chloride is:

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

  1. $68$%

  2. $34$%

  3. $48$%

  4. $17$%

Show answer
Correct Option: B
Explanation:

$(NH _3)Cl+NaOH\rightarrow +NH _3\uparrow +H _2O$

excess $H _2SO _4$ reg. $20ml$ of $1N\,\,NaOH$
$\Rightarrow $ moles of $H _2SO _4=(20\times 1)milimoles$
                                 $=20m\,mol$
added amount of $H _2SO _4=\dfrac{N}{5}\times 150ml=30mmol$
amount of $h _2so _4$ reacted with $nh _3=0.01MOLE$
$\Rightarrow 0.01\, mole$ of $NH _3$ present in $(NH _4)Cl$
So, $\%purity=\dfrac{(0.01)\times 17}{0.5}\times 100=34\%$

A sample of $CaCO _3$ is 50% pure. On heating $1.12 L$ of $CO _2$ (at STP) is obtained. Residue left (assuming non-volatile impurity) is

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

  1. 7.8 g

  2. 3.8 g

  3. 2.8 g

  4. 8.9 g

Show answer
Correct Option: A
Explanation:

Solution:- (A) $7.8 \; g$

Volume of $C{O} _{2}$ formed $= 1.12 \; L$
At STP, volume of $1$ mole of gas $= 22.4 \; L$
$\therefore$ No. of moles of $C{O} _{2}$ formed $= \cfrac{1.12}{22.4} = 0.05 \text{ mol}$
Heating of $CaC{O} _{3}$-

$CaC{O} _{3} \longrightarrow CaO + C{O} _{2}$
From the above reaction,
$1$ mole of $C{O} _{2}$ is formed on heating $1$ mole of $CaC{O} _{3}$.
Therefore,
$0.05$ mole of $C{O} _{2}$ is formed on heating $0.05$ mole of $CaC{O} _{3}$.

Molecular weight of $CaC{O} _{3} = 100 \; g$
$\therefore$ Weight of $CaC{O} _{3} = 0.05 \times 100 = 5 \; g$
As the sample was $50 \%$ pure.
Thus the $50 \%$ of the sample was heated in the form of $CaC{O} _{3}$.
Amount of sample left unreacted $= 5 \; g$
Also,
No. of moles of $CaO$ formed $= 0.05$
Molecular weight of $CaO = 56 \; g$
Weight of $CaO = 56 \times 0.05 = 2.8 \; g$
Therefore,
Amount of residue left $= 5 + 2.8 = 7.8 \; g$

In ayurvedic preparation of swarnabhasma, what purity of golden will be used?

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

  1. $42$ % gold + silver

  2. $91$ % amalgam

  3. $24$ carat

  4. $58.5$% copper

Show answer
Correct Option: A
Explanation:

Swarna Bhasma is prepared from Gold. It is used in Ayurvedic treatment of infertility, asthma, tissue wasting, poisoning etc. This medicine should only be taken strictly under medical supervision.

In the given question option A is the correct answer.

One mole of photons is known as one Einstein of radiation. According to Stark-Einstein law of photochemical equivalence, one mole of reactant absorbs one Einstien of energy. For a photochemical reaction, a term called quantum yield is defined as:
Quantum yield $ (\phi) = \dfrac {No. \,of \,moles \,of \,reactant \,converted} {No. \,of \,Einstein \,absorbed} $
The correct statement(s) is/are:

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

  1. for a chain reaction $\phi _{gas} >> \phi _{solution}$

  2. in a photochemical chain reaction $\phi >> 1$

  3. in a photochemical chain reaction $\phi << 1$

  4. for a chain reaction $\phi _{gas} << \phi _{solution}$

Show answer
Correct Option: A,B
Explanation:
$Quantum \ Yield= \cfrac {Number \ of \ moles \ of \ reactant \ converted}{Number \ of \ einstein \ absorbed}$
An einstein of radiation $=$ one mole of photons
For a chain reaction, $\phi _{gas} >> \phi _{solution}$
and ln photochemical chain reaction $\phi >>1$

A sample of $CaC{O _3}$ is $50\% $ pure. On heating $1.12{\text{ }}L$ of $C{O _2}$ (at STP) is obtained. Residue left (assuming non-volatile impurity) is:

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

  1. 7.8 g

  2. 3.8 g

  3. 2.8 g

  4. 8.9 g

Show answer
Correct Option: A
Explanation:

No. of moles of $CO _2$ evolved $=\cfrac{1.12}{22.4}=0.05$ $moles$

$CaCO _3(s)\overset { \Delta }{ \longrightarrow } CaO\downarrow+CO _2\uparrow$
                             $0.05$        $0.05$ $moles$
So, $0.05$ $moles$ of $CaCO _3$ have  reacted.
Mass $=0.05\times 100=5$ $gm=50\%$ of $CaCO _3$ sample
Total weight $=2\times 5$ $gm=10$ $gm$
Residue left by $CaCO _3=5$ $gm$
Residue left by $CaO=56\times 0.05=2.8$ $gm$
Toatl residue $=5+2.8=7.8$ $gm$