Q: When 22.4 litres of H2( g) is mixed with 11.2 litres of Cl2( g) , each at S. T.P., the moles of HCl(g) formed is equal to :

nH2=22.4 LV(L)=22.422.4=1 nC2=22.411.2=0.5 mole after reaction H2( g)+Cl2( g)→2HCl(g)1 mole initially - 0.5 mole(1−0.5)00.5×2=0.5 mole 0=1 mole

Q: What amount of carbon will be required to form 8.8 g of CO2 after complete combustion :

C+O2(g)→CO2(g) Moles of CO2=448.8 =0.2 Moles of carbon = moles of CO2=0.2 Mass of carbon=0.2×12=2.4 g

Q: 25 g of an unknown hydrocarbon upon burning produces 88 g of CO2 and 9 g of H2O. This unknown hydrocarbon contains

Hydrocarbon containing C and H upon burning produces CO2 and water vapour respectively. The equation is represented as CxHy+(x+y/4)O2⟶xCO2+(y/2)H2O Mass of carbon =4412× mass of CO2 =4412×88 g=24 g Mass of hydrogen =182× mass of H2O =182×9=1 g So, the unknown hydrocarbon contains 24 g of carbon and 1 g of hydrogen.

Q: The mass of potassium dichromate crystals required to oxidise 750 cm3 of 0.6M Mohr's salt solution is (molar mass =392 )

Mohr's salt is FeSO4⋅(NH4)2SO4⋅6H2O. Only oxidisable part Fe2+ is [Fe2+⟶Fe3++e−]×6 Cr2O72−+14H++6e−⟶2Cr3++7H2O 6Fe2++Cr2O72−+14H+⟶ 6Fe3++2Cr3++7H2O Millimoles of Fe2+=750×0.6=450 Moles of Fe2+=1000450=0.450 mol 6 molFeFel2+≡1 molCr2O72− ∴0.450 molFe2+≡60.450=0.075 molCr2O72− =0.075×294 g=22.05 g

Q: SO2 may be used for production of CS2 by using its reaction with coke. The products are CO and CS2. Using one kg of SO2 with an efficiency of conversion of 80%, calculate the weight of CS2 produced in grams.

(a) 475 g2SO2+5C→CS2+4CO2 moles of SO2 produce 1 mole CS2

Q: Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is

(d) The following reaction occurs: 6Fe2++Cr2O72−+14H+→6Fe3++2Cr3++7H2O From the above equation, we find that Mohr's salt (FeSO4⋅(NH4)2SO4⋅6H2O) and dichromatic reacts in 6:1 molar ratio.

Q: Propane burns in air according to the following equation: C3H8+5O2→3CO2+4H2O What volume of propane is consumed on using 1000 cm3 of air, considering only 20% of air contains oxygen?

Propane burns in air according to the following equation:C3H8+5O2→3CO2+4H2O1 mole of propane combines with 5 moles of oxygen.Volume is directly proportional to the number of moles.1 cm3 of propane will combine with 5 cm3 of oxygen.Only 20% of air contains oxygen. So 1000 cm3 of air contains 1000 cm3×10020= 200 cm3 of oxygen.The volume of propane that is consumed on using 1000 cm3 of air is5200 cm3=40 cm3.

Question 1

When 22.4  litres of H2​( g)  is mixed with 11.2  litres of Cl2​( g) , each at S.  T.P., the moles of HCl(g) formed is equal to :

Accepted Answer

nH2​​=22.4 LV(L)​=22.422.4​=1 nC2​​=22.411.2​=0.5 mole  after reaction ​H2( g)​+Cl2( g)​→2HCl(g)​1 mole initially - ​0.5 mole(1−0.5)​00.5×2​=0.5 mole 0​=1 mole ​

Question 2

For the given reaction
2A+3B→4C
How many moles of C will be formed by reacting 10 moles of A and 24 moles of B :

Accepted Answer

Here, the limiting reagent is A,Hence, moles of C=4× 2moles of A​ =24×10​=20 moles

Question 3

What amount of carbon will be required to form 8.8 g of CO2​ after complete combustion :

Accepted Answer

C+O2​(g)→CO2​(g)   Moles of CO2​=448.8​ =0.2  Moles of carbon = moles of CO2​=0.2 Mass of carbon=0.2×12=2.4 g

Question 4

25 g of an unknown hydrocarbon upon burning produces 88 g of CO2​ and 9 g of H2​O. This unknown hydrocarbon contains

Accepted Answer

Hydrocarbon containing C and H upon burning produces CO2​ and water vapour respectively. The equation is represented as

Cx​Hy​+(x+y/4)O2​⟶xCO2​+(y/2)H2​O
 Mass of carbon =4412​× mass of CO2​
=4412​×88 g=24 g
 Mass of hydrogen =182​× mass of H2​O
=182​×9=1 g

So, the unknown hydrocarbon contains 24 g of carbon and 1 g of hydrogen.

Question 5

The mass of potassium dichromate crystals required to oxidise 750 cm3 of 0.6M Mohr's salt solution is (molar mass =392 )

Accepted Answer

Mohr's salt is FeSO4​⋅(NH4​)2​SO4​⋅6H2​O.
Only oxidisable part Fe2+ is

[Fe2+⟶Fe3++e−]×6

Cr2​O72−​+14H++6e−⟶2Cr3++7H2​O
6Fe2++Cr2​O72−​+14H+⟶
6Fe3++2Cr3++7H2​O
 Millimoles of Fe2+=750×0.6=450
 Moles of Fe2+=1000450​=0.450 mol
6 molFeFel2+≡1 molCr2​O72−​
∴0.450 molFe2+≡60.450​=0.075 molCr2​O72−​
=0.075×294 g=22.05 g

Question 6

SO2​ may be used for production of CS2​ by using its reaction with coke. The products are CO and CS2​. Using one kg of SO2​ with an efficiency of conversion of 80%, calculate the weight of CS2​ produced in grams.

Accepted Answer

(a) 475 g2SO2​+5C→CS2​+4CO2 moles of SO2​ produce 1 mole CS2​

Question 7

Analysis of dissolved oxygen in river water is very essential to ensure that aquatic life is safe. As per European standards, dissolved oxygen should never be less than 6.4mg/L and preferably should be greater than 9.6mg/lit. A water body having value less than 6.4mg/L is considered 'heavily polluted' and a water body having value greater than 9.6mg/L is considered optimum fit for aquatic life. One method to measure dissolved oxygen is through Winkler titration. The method involves three stages:
Stage 1: The water sample is treated with excess of Mn2+ (aq) solution along with a base to cause the following reaction:
Mn2+(aq)+2OH−(aq)→Mn(OH)2​( s)
The precipitate then reacts with dissolved oxygen as shown below.
2Mn(OH)2​( s)+21​O2​(aq)+H2​O(l)→2Mn(OH)3​ (s) ...(ii) 
Stage 2: The solution is added with excess of acidified solution of KI to cause the following reaction:
2Mn(OH)3​( s)+2KI+3H2​SO4​
→2MnSO4​+K2​SO4​+I2​(aq)+6H2​O
Using the above information answer the following questions:
1 If 1 L of water sample is taken then calculate the maximum sample is considered 'heavily polluted'.
 (a) 80 mL​ (b) 80 L​ (c) 120 mL​ (d) 120 L​
2 If 1 L of water sample is taken then calculate maximum volume of 0.1MNa2​ S2​O3​ solution required if the sample is considered 'optimum fit' for aquatic life.
(a) 8 mL
(b) 80 mL
(c) 12 mL
(d) 120 mL
3. Which of the following statements is correct w.r.t. above method?
(a) Greater the amount of dissolved oxygen, lesser will be moles of Na2​ S2​O3​ required.
(b) Excess solution in Stage 1 ensure all dissolved oxygen reacts,
(c) Moles of Mn(OH)2​( s) reacted and moles of Mn(OH)3​ formed will be different.
(d) 1 mole of dissolved oxygen will require 2 moles of Na2​ S2​O3​ for estimation.

Accepted Answer

1(a)80 ml2 (c) 12 mL3. (b) Excess solution in Stage 1 ensure all dissolved oxygen reacts,

Question 8

Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is

Accepted Answer

(d) The following reaction occurs:
6Fe2++Cr2​O72−​+14H+→6Fe3++2Cr3++7H2​O
From the above equation, we find that Mohr's salt (FeSO4​⋅(NH4​)2​SO4​⋅6H2​O) and dichromatic reacts in 6:1 molar ratio.

Question 9

Two flasks A and B of 500 mL each are respectively filled with O2​ and SO2​ at 300 K and 1 atmospheric pressure. The flasks will contain:

Accepted Answer

According to Avogadro's Hypothesis: Equal volumes of all gases contain equal number of atoms under similar conditions of temperature and pressure.

Question 10

Propane burns in air according to the following equation: C3​H8​+5O2​→3CO2​+4H2​O What volume of propane is consumed on using 1000 cm3 of air, considering only 20% of air contains oxygen?

Accepted Answer

Propane burns in air according to the following equation:C3​H8​+5O2​→3CO2​+4H2​O1 mole of propane combines with 5 moles of oxygen.Volume is directly proportional to the number of moles.1 cm3 of propane will combine with 5 cm3 of oxygen.Only 20% of air contains oxygen. So 1000 cm3 of air contains 1000 cm3×10020​= 200 cm3 of oxygen.The volume of propane that is consumed on using 1000 cm3 of air is5200 cm3​=40 cm3.

Question 11

What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?

Accepted Answer

C2​H4​+3O2​⟶2CO2​+2H2​O 28 kg96 kg Weight of oxygen required for the Complete combustion of 28 kg ethylene = 96 kg so,Weight of oxygen required for the combustion of 2.8 kg ethylene.W=2896×2.8​=9.6 kg

Question 12

An amount of 5 millimoles of LiAlH4​ was treated with 20 millimoles of t-butylalcohol. A total of 15 millimoles of hydrogen was evolved for the following reaction.
LiAlH4​+3(CH3​)3​COH→Li[(CH3​)3​CO]3​AlH+3H2​
The addition of an excess of another alcohol, methanol, to the above reaction mixture caused the fourth H atom of the LiAlH4​ to be replaced according to the following equation.
Li[(CH3​)3​CO]3​AlH+CH3​OH→Li[(CH3​)3​CO]3​(CH3​O)Al+H2​

How many millimoles of H2​ was evolved due to the addition of CH3​OH ?

Accepted Answer

ReactionInitialmole​LiAlH4​5mmol​+3(CH3​)3​COH20mmol​→​Li[(CH3​)3​CO]3​5mmol​+3H2​15mmol​​Excess of alcoholReaction​Li[(CH3​)3​CO]3​5mmol​+CH3​OHexcess​​→​Li[(CH3​)3​CO3​][CH3​O]Al5mmol​+H2​​​So, 5 mmol of hydrogen was evolved due to the addition of CH3​OH.

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