Find out the volume of Cl 2 at STP produced by the action of 100 cm 3 of 0 . 2 N HCl on excess of MnO 2 .

The balanced reaction between HCl and MnO 2 is following, MnO 2 + 4 HCl 4 × 36 . 5 → MnCl 2 + 2 H 2 O + Cl 2 22400 cm 3 at STP Normality = weight of solute g × 1000 volume of solvent × Eq . Wt . of solute N = W × 1000 V mL × Eq . Wt ⋯ ⋯ ⋯ ⋯ ⋯ ( i ) And Eq . Wt . = Molar mass n f ⋯ ⋯ ⋯ ⋯ ⋯ ( ii ) Where n f is number of H + or OH - replaced by one mole of acid or base in a reaction. From the equation i and ii , 0 . 2 = W × 1000 100 × 36 . 5 1 W = 0 . 2 × 100 × 36 . 5 1000 W = 0 . 73 g ∵ 4 × 36 . 5 g of HCl produced = 22400 cm 3 of Cl 2 at STP ∴ 0 . 73 g of HCl will produce = 22400 4 × 36 . 5 × 0 . 73 cm 3 = 112 cm 3 of Cl 2 at STP

Zinc and hydrochloric acid react according to the reaction: Zn ( s ) + 2 HCl ( aq ) → ZnCl 2 ( aq ) + H 2 ( g ) If 0 . 30 mol Zn are added to hydrochloric acid containing 0 . 52 mol of HCl , how many moles of H 2 are produced?

Zinc reacts with hydrochloric acid by the given reaction, Zn ( s ) 1 mol + 2 HCl ( aq ) 2 mol → ZnCl 2 ( aq ) + H 2 1 mol ( g ) Let Zn s completely consumed in the reaction. ∵ 1 mol Zn produces = 1 mol H 2 ∴ 0 . 30 mol Zn will produce = 1 1 × 0 . 30 = 0 . 3 mol H 2 Let HCl completely consumed in the reaction. ∵ 2 mol HCl produces = 1 mol H 2 ∴ 0 . 52 mol HCl will produce = 1 2 × 0 . 52 = 0 . 26 mol H 2 From the above, the HCl gives the least amount of hydrogen gas . Therefore, HCl is a limiting reagent and 0 . 26 mol of hydrogen gas will be produced.

500 cm 3 of 0 . 250 M Na 2 SO 4 solution added to an aqueous solution of 15 . 00 g of BaCl 2 resulted in the formation of a white precipitate of BaSO 4 . How many moles and how many grams of BaSO 4 are formed?

The following reaction occurs when Na 2 SO 4 added to an aqueous solution of BaCl 2 , Na 2 SO 4 1 mol + BaCl 2 208 . 23 g 1 mol → BaSO 4 1 mol + 2 NaCl ⋯ ⋯ ⋯ ⋯ ( i ) As we know that molarity = Number of moles of solute volume of solution L ∴ moles = molarity × volume L Moles of Na 2 SO 4 = 0 . 250 × 500 × 10 - 3 = 0 . 125 mol And the moles of BaCl 2 = Mass of BaCl 2 Molar mass of BaCl 2 moles of BaCl 2 = 15 208 . 23 = 0 . 0720 mole From the above, BaCl 2 is a limiting reagent and the BaSO 4 forms according to BaCl 2 . ∵ 1 mol of BaCl 2 produces = 1 mol BaSO 4 ∴ 0 . 0720 mol of BaCl 2 will produce = 1 1 × 0 . 0720 = 0 . 0720 mol BaSO 4 And the amount of BaSO 4 produces = Moles of BaSO 4 × molar mass of BaSO 4 = 0 . 0720 × 233 . 38 = 16 . 80 g

HARD

11th CBSE

IMPORTANT

Earn 100

What volume of a solution of $HCl$ containing $73$ g/litre would suffice for the exact neutralization of $NaOH$ obtained by allowing $0.46 g$ of metallic sodium to react with water.

Important Questions on Some Basic Concepts of Chemistry

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 20

Find out the volume of

{Cl}_{2}

at STP produced by the action of

100 {cm}^{3}

0.2 N HCl

on excess of

1000 mL

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 21

A mixture of ethane

(C_{2} H_{6})

and ethene

(C_{2} H_{4})

occupies

40

litres at

1.00

atm and at

400

K. The mixture reacts completely with

130

g of

O_{2}

to produce

{CO}_{2}

and

H_{2} O

. Assuming ideal gas behavior, calculate the mole fractions of

C_{2} H_{4}

and

C_{2} H_{6}

in the mixture.

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 1

5.0

g of marble was added to

7.5

g dilute hydrochloric acid. After the reaction was over, it was found that

0.5

g of marble was left unused. Calculate the percentage strength of hydrochloric acid. What volume of

{CO}_{2}

measured at STP will be evolved in the above reaction?

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 2

Calculate the volume of

1.00

mol L^-1 aqueous sodium hydroxide solution that is neutralized by

200

mL of

2.00

mol L^–1 aqueous hydrochloric acid and mass of sodium chloride produced.

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 3

Bromine is prepared commercially by the reaction:

2 {Br}^{-} (aq) + {Cl}_{2} (aq) \to 2 {Cl}^{-} (aq) + {Br}_{2} (aq)

Suppose we have

50.0

mL of

0.060

M solution of

NaBr

. What volume of

0.050

M solution of

{Cl}_{2}

is needed to react completely with the

{Br}^{-}

?

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 1

20.0

g of

{CaCO}_{3}

is treated with

20.0

g of

HCl

, how many grams of

{CO}_{2}

will be produced?

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 2

Zinc and hydrochloric acid react according to the reaction:

Zn (s) + 2 HCl (aq) \to {ZnCl}_{2} (aq) + H_{2} (g)

0.30

mol

Zn

are added to hydrochloric acid containing

0.52

mol of

HCl

, how many moles of

H_{2}

are produced?

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 3

500

{cm}^{3}

0.250 M {Na}_{2} {SO}_{4}

solution added to an aqueous solution of

15.00 g

{BaCl}_{2}

resulted in the formation of a white precipitate of

{BaSO}_{4}

. How many moles and how many grams of

{BaSO}_{4}

are formed?

What volume of a solution of HCl containing 73 g/litre would suffice for the exact neutralization of NaOH obtained by allowing 0.46 g of metallic sodium to react with water.

Important Questions on Some Basic Concepts of Chemistry

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What volume of a solution of $HCl$ containing $73$ g/litre would suffice for the exact neutralization of $NaOH$ obtained by allowing $0.46 g$ of metallic sodium to react with water.