Chemistry

Mole Concept and Stoichiometry

365 Questions

The mole concept is a core area of chemistry focusing on the quantitative relationships between reactants and products in chemical reactions. Questions cover molecular mass calculations, stoichiometric conversions, and determining the number of molecules. This topic is essential for most science entrance and competitive examinations.

molecular mass calculationsstoichiometric conversionsmole fraction problemsheavy water reactionsoxidation calculations

Mole Concept and Stoichiometry Questions

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\%$

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

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?

  1. $26$%
  2. $38$%
  3. $52$%
  4. $76$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$26$ ml of $NH _3$ yeild into 100 ml 

$=26/100$ 
= $26%$

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

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?

  1. $-10$%
  2. $+90$%
  3. $+10$%
  4. $+111$%
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $= \dfrac{1.8}{2} \times 100 = 90\%$

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

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:

  1. $12$%
  2. $98$%
  3. $73$%
  4. $43$%
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  1. $80$
  2. $70$
  3. $60$
  4. $50$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 %

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

  1. 36.8 g

  2. 62.5 g

  3. 80 g

  4. 91.98 g
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $(D)$.

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

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?

  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 $H _2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

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

  1. 7.8 g

  2. 3.8 g

  3. 2.8 g

  4. 8.9 g
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $= 5 + 2.8 = 7.8 \; g$

Multiple choice zoology excretion in living organisms excretion - animals human excretory system and products excretory system and products excretion - plants excretion in plants

How many molecules of ammonia are required to form 8 molecules of urea?

  1. 24

  2. 8

  3. 16

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

Urea synthesis is a five-step cyclic process, with five distinctive enzymes. One turn of the cycle consumes 2 molecules of ammonia and  1 molecule of carbon dioxide and creates 1 molecule of urea ((NH2)2CO. Hence  16  molecules of ammonia are required to form 8 molecules of urea.

So, the correct answer is 'Option C'.

Multiple choice maths fraction one fraction, many forms use of decimals in measure of length decimal fractions in the units of currency, length, weight, capacity

How much pure alcohol should be added to $400 ml$ of strength $15\%$ to make its strength $32\%$?

  1. $50 ml$
  2. $75 ml$
  3. $100 ml$
  4. $150 ml$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The amount of alcohol in $400ml$ solution of $15\%$ strength

$=400\times \cfrac { 15 }{ 100 } ml=60ml$

Let the required amount of alcohol to be added $=x$ ml.
It will increase the amount of the solution by $x$ ml.

$\therefore $ The amount of alcohol now $=(60+x)ml$
and the amount of the solution$=(400+x)ml$.
Then, the strength $=32\%=\cfrac { 60+x }{ 400+x } =\cfrac { 32 }{ 100 }$

$\Rightarrow 25x+1500=3200+8x\ \Rightarrow 17x=1700\ \Rightarrow x=100$.
So, the required amount of pure alcohol to be added $=100 ml$.