Hydrogen Test

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Hydrogen Test - 56

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Weekly Quiz Competition

Question 1
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Which of the following statement(s) is/are correct about the water-gas shift reaction which is used in the commercial preparation of hydrogen gas?

A
It involves the selective oxidation of $$CO$$ of water gas to $$CO_{2}$$ with the help of steam in the presence of iron chromate as catalyst

B
It involves the separation of $$CO$$ and $$H_{2}$$ by solvent extraction method

C
It involves the absorption of $$CO$$ in aqueous copper $$(I)$$ chloride solution

D
It involves the removal of $$H_{2}$$ through occlusion with $$Pd$$

Solution

$$CO_{(g)} + H_{2(g)} + H_2 O_{(g)} \xrightarrow [673\: K] {FeCrO_4} 2 H_{2(g)} + CO_{2(g) }$$

Water-gas shift reaction involves the selective oxidation of $$CO$$ of water gas to $$CO_2$$ with the help of steam, in the presence of iron chromate as catalyst.
Question 2
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Half litre each of three samples of $$H_{2}O_{2}$$ labelled $$10$$ volume, $$15$$ volume and $$20$$ volume are mixed and then diluted with equal volume of water. The relative strength of resultant $$H_{2}O_{2}$$ solution is:

A
$$2.5$$

B
$$5.0$$

C
$$7.5$$

D
$$10$$

Solution

Volume strength of $$H_{ 2 }{ O }_{ 2 } = 5.6\times \text{N}$$ $$(\text{N} =$$ Normality$$)$$

Normality of $$10$$ vol of $$H_{ 2 }{ O }_{ 2 } = \dfrac{10}{5.6}\text{N}$$

Normality of $$15$$ vol of $$H_{ 2 }{ O }_{ 2 } = \dfrac{15}{5.6} \text{N}$$

Normality of $$20$$ vol of $$H_{ 2 }{ O }_{ 2 } = \dfrac{20}{5.6} \text{N}$$

Let $$500$$ ml of each is mixed, then the total volume of mix $$= 1500$$ ml.

Also, this is diluted to double its volume, i.e., $$= 3000$$ ml

$$\text{N}\times 3000$$ = $$\dfrac { 10 }{ 5.6 } \times 500+\dfrac { 15 }{ 5.6 } \times 500+\dfrac { 20 }{ 5.6 } \times 500$$

Normality of $$H_{ 2 }{ O }_{ 2 }$$ mix $$= 1.339\:\text{N}$$

Vol. strength of $$H_{ 2 }{ O }_{ 2 } = 1.339 \times 5.6 = 7.5$$.
Question 3
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If $$10$$ ml of $$H_2O_2$$ solution requires $$25$$ ml of $$\dfrac{\text{N}}{20}\: KMnO_{4}$$ for complete oxidation in acidic medium, then the mass (in g) of available $$O_{2}$$ per litre present in the solution is:

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

The oxidation reaction is as follows:
$$2KMn{ O }_{ 4 } + 5{ H }_{ 2 }{ O }_{ 2 }+ 4{ H }_{ 2 }{ SO }_{ 4 }\rightarrow 2{ KHSO }_{ 4 }+ 2MnS{ O }_{ 4 }+ 8{ H }_{ 2 }O+ 5{ O }_{ 2 }$$
Meq. of $${ O }_{ 2 } =$$ Meq. of $${ H }_{ 2 }{ O }_{ 2 } =$$ Meq. of $$KMn{ O }_{ 4 }$$
It is given that $$10$$ ml of $$H_2O_2$$ solution when treated with $$25$$ ml of $$\dfrac {N}{20}\ KMnO_4$$ for complete oxidation.
$$\text{N} \times 10 = 25 \times \dfrac{1}{20}$$
Normality of $${ H }_{ 2 }{ O }_{ 2 } = 0.125\:\text{N} = $$Normality of $${ O }_{ 2 }$$
Weight of $${ O }_{ 2 }= 0.125 \times 32 = 4$$ g per litre.
Question 4
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$$25$$ ml of $$H_{2}O_{2}$$ solution were added to excess of acidified $$KI$$ solution. The iodine so liberated required $$20$$ ml of $$0.1\: \text{N}$$ $$Na_{2}S_{2}O_{3}$$ for titration. Calculate strength of $$H_2O_2$$ in terms of normality and percentage?

A
$$0.04\:\text{N}, 0.136\%$$

B
$$0.08\:\text{N}, 0.136\%$$

C
$$0.04\:\text{N}, 0.163\%$$

D
$$0.02\:\text{N}, 0.163\%$$

Solution

$$I_{2}+2Na_{2}S_{2}O_{3}^-\rightarrow Na_{2}S_{4}O_{6}+2NaI$$

Milliequivalent of $$H_2O_2 = $$ Milliequivalent of $$I_2 = $$Milliequivalent of $$Na_2S_2O_3$$

$$\dfrac { w }{ 34/2 } \times 1000 = 20 \times 0.1$$

$${ \therefore W }_{ H_{ 2 }{ O }_{ 2 } } = 0.034$$ gm$$/ 25$$ ml

$${ N }_{ H_{ 2 }{ O }_{ 2 } } = \dfrac { 0.034 }{ 34/2 } \times \dfrac { 1000 }{ 25 } = 0.08$$

Volume strength $$= 5.6\times0.08 = 0.448$$

$$\%$$ strength $$= \dfrac { 17 }{ 56 } \times 5.6 \times 0.08 = 0.136%$$.
Question 5
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Peroxoborate in aqueous solution provides:

A
$$[B(OH)_4]^-+H_2O_2$$

B
$$Na_2B_4O_8\cdot 10H_2O$$

C
$$HBO_2$$

D
$$Na[B_3O_3(OH)_4]$$

Solution

Hence, peroxoborate, $$B_2O_4(OH)_4$$ in aqueous solution provides peroxide, $$H_2O_2$$ and $${B(OH)_4}^-$$.
Question 6
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An experimental set up is shown in the figure. Identify the metal X and gas Y respectively in the figure.

A
Zinc, $$\displaystyle H_{2}$$ gas

B
Copper, $$\displaystyle CO_{2}$$ gas

C
Silver, $$\displaystyle CO_{2}$$ gas

D
Platinum, $$\displaystyle H_{2}$$ gas

Solution

Zinc reacts with dil. HCl to give hydrogen gas which is collected above water as it is the lightest gas while other metals are unreactive towards dil. HCl as they are inert present below the hydrogen in the reactivity series.

$$Zn(s) +HCl(dil.)\rightarrow ZnCl_2(aq) +H_2(g)$$

Option A is correct.
Question 7
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When a substance A reacts with water, it produces a combustible gas B and a solution of substance C in water. D reacts with this solution of C and produces the same gas B on warming. D can also produce gas B on reaction with dilute $$ \displaystyle H_{2}SO_{4}$$. A imparts a deep golden yellow colour to smokeless flame. A, B, C and D respectively are:

A
$$Na \displaystyle, H_{2}, NaOH, Zn$$

B
$$K, \displaystyle H_{2}, KOH, Al$$

C
$$Ca, \displaystyle H_{2}, \displaystyle Ca\left ( OH \right )_{2}, Sn$$

D
$$CaC_{2},\displaystyle C_{2}H_{2},\displaystyle Ca\left ( OH \right )_{2}, Fe$$

Solution

Reactions:

$$2Na+2H_2O\rightarrow 2NaOH+H_2$$
A C B

$$Na$$ gives a golden colour flame.

$$Zn+2NaOH(dil.)\rightarrow Na_2ZnO_2+H_2\uparrow$$
D C B
Question 8
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A teacher one day pointed out to his students the peculiar fact that water is a unique liquid that freezes exactly at $$0^oC$$ and boils exactly at $$100^oC$$. He asked the students to find the correct statement based on this fact.

Which of the statements is true regarding the above fact?

A
Water dissolves anything however sparingly the dissolution may be.

B
Water is a polar molecule.

C
Boiling and freezing temperatures of water were used to define a temperature scale.

D
Liquid water is denser than ice.

Solution

Option (C) is correct.
Boiling and freezing temperatures of water were used to define a temperature scale.The scaling of zero degree centigrade is based on this fact.
Question 9
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By which process $$H_{2}O_{2}$$ does not form?

A

B

C

D
By $$(A)$$ and $$(C)$$ process

Solution

Hydrogen peroxide is prepared by Merck's method in which sodium peroxide reacts with dilute acids.

$$Na_2O_2+H_2SO_4 \rightarrow Na_2SO_4 + H_2O_2$$.

While $$KO_2$$ and $$PbO_2$$ does not form hydrogen peroxide on reacting with dilute acids.

Option D is correct.
Question 10
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Which of the following produces hydrolith with dihydrogen?

A
Mg

B
Al

C
Cu

D
Ca

Solution

Calcium reacts with dihydrogen to produce hydrolith which is calcium hydride.

$$Ca+H_2 \xrightarrow {\Delta} CaH_2$$