Tag: chemistry

Questions Related to chemistry

When the fertilizer plant completes its small batch process, they find that they have only collected $26\ mL$ of $NH _{3}$ which was supposed to be $100\ mL$. They are disappointed with this result due to the low yield. What is the % yield of their process?

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

  1. $26$%

  2. $38$%

  3. $52$%

  4. $76$%

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Correct Option: A
Explanation:

$26$ ml of $NH _3$ yeild into 100 ml 

$=26/100$ 
= $26%$

A reaction is known to have a percent yield of $80$%. It the actual yield is $25$ grams, what was the expected yield?

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

  1. $3.2\ grams$

  2. $19\ grams$

  3. $31.25\ grams$

  4. $190\ grams$

Show answer
Correct Option: C
Explanation:

Percent yield is the ratio of actual yield to the expected yield of compound.

$\%$ Yield $= \dfrac{\text{Actual Yield}}{\text{Expected Yield}} \times 100$
$\Rightarrow 80 = \dfrac{25}{\text{Expected Yield}} \times 100$
$\Rightarrow$ Expected Yield $= \dfrac{25}{80} \times 100 = \dfrac{250}{8} = 31.259\space g$

A bacterium can convert xylose to lactic acid with a percent yield of $85$%. What mass of lactic acid would be produced from $50.0\ g$ sample of xylose by this bacterium? (At. mass of xylose$(C _5H _{10}O _5) =150\ g/mol,$ At. of lactic acid$(C _3H _6O _3)=90\ g/mol$).

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

  1. $58.8\ g$

  2. $85\ g$

  3. $50\ g$

  4. $42.5\ g$

Show answer
Correct Option: D
Explanation:

Percent Yield is the ratio of actual yield of a substance to the theoretical yield.

Percent Yield $= \dfrac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$
Given, Theoretical Yield $50\space g,$ Percent Yield $= 85$
So, $(\dfrac{85}{100} \times 50) =$ Actual Yield
$\Rightarrow$ Actual Yield $= 42.5 \space g$

A student conducts an experiment to produce a $Ca{CO} _{3}$ precipitate. The student collects $1.80\ g$ of product after predicting it should be possible to produce $2.00\ g$ of the product.
What is the student's percent yield for this experiment?

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

  1. $-10$%

  2. $+90$%

  3. $+10$%

  4. $+111$%

Show answer
Correct Option: B
Explanation:

Percent Yield is the ratio of actual yield to the theoretical yield.

$\Rightarrow \% $ Yield $= \dfrac{E}{T} \times 100$
$\Rightarrow$ Here, $E = 1.8 \space g; \space T = 2\space g$
So, percent yield $= \dfrac{1.8}{2} \times 100 = 90\%$

If $100\ mL$ of the acid is neutralised by $100\ mL$ of $4\ M\ NaOH$, the purity of concentrated $HCl$ (sp. gravity $= 1.2)$ is:

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

  1. $12$%

  2. $98$%

  3. $73$%

  4. $43$%

Show answer
Correct Option: A
Explanation:

No of equivalent of acid $=$ No of equivalent of base.

$100 \times m _1=100 \times4$
$m _1=4m$
$4$mole of $HCL$ in $1$liter of water..........$(1)$
$\frac{{\rho HCl}}{{\rho {H _2}O}} = k$
density of $HCl=1200gm/l$...........$(2)$
purity of $HCl\Rightarrow 1200gm\,\,in\,1liter$
purity of $HCl$ in gram in $1$ liter $\Rightarrow 1200gm$
purity of $HCl \Rightarrow$ $\frac{{4 \times 36.5}}{{1200}} \times 100 \Rightarrow  \sim 12\% $
hence, purity of $HCl$ is $ \sim 12\% $
so option $A$ is correct.

$50\ g$ of a sample of $NaOH$ required for complete neutralisation of $1\ litre\ N\ HCl$. What is the percentage purity of $NaOH$?

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

  1. $80$

  2. $70$

  3. $60$

  4. $50$

Show answer
Correct Option: A
Explanation:

for complete neutralisation we have

moles of acid = moles of base
moles of acid = normality $$ volume = $1 *1$
moles of base = 1 =$\dfrac{given mass}{molecular mass}$

so, given mass = moles $$ molecular mass = 1$*$ = 40

percentage purity = $\dfrac{40 }{50} * 100$ = 80 %

In a reaction vessel,100 g $H _2$ and 100 g $Cl _2$ are inbred and suitable conditions are provided for take following reaction: 

$H _2 ( g)+ Cl _2(g)\rightarrow 2HCl(g)$


The amount of $HCI$ formed (at 90% yield) will be:

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

  1. 36.8 g

  2. 62.5 g

  3. 80 g

  4. 91.98 g

Show answer
Correct Option: D
Explanation:

100 g of $H _2$ = 50 mole

100 g of $Cl _2$ = 1.4 mole
According to the reaction 1 mole of $H _2$ is reacting with 1  mole of $Cl _2$
So, $Cl _2$ is th limiting reagent 
So, moles of $HCl$ formed = 2 $\times$ 1.4 mole
 
Amount of $HCl$ = 2.8 $\times$ 36.5 = 102.2 g

Since, the reaction is giving 90 % yield
therefore , amount of $HCl$ formed = $102.2 \times \dfrac{90}{100}$

Amount of HCl formed = 91.98 g

Hence, the correct option is $(D)$.

In the Haber process:
$N _2(g) + 3H _2(g) \rightarrow 2NH _3(g) $
$30 L$ of $H _2$ and $30 L$ of ${N _2}{^-}$ were taken for reaction which yielded only $50\%$ of expected product. What will be the composition of the gaseous mixture in the end?

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

  1. 20 L $NH _3$, 25 L $N _2$ and 20 L $H _2$

  2. 10 L $NH _3$, 25 L $N _2$ and 15 L $H _2$

  3. 20 L $NH _3$, 10 L $N _2$ and 30 L $H _2$

  4. 20 L $NH _3$, 25 L $N _2$ and 15 L $H _2$

Show answer
Correct Option: A

$0.5\ g$ 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 $= 5 + 2.8 = 7.8 \; g$