Ethane burns in oxygen to form CO 2 and H 2 O according to the equation 2 C 2 H 6 + 7 O 2 → 4 CO 2 + 6 H 2 O . If 1250 cc of oxygen is burnt with 300 cc of ethane. Calculate the volume of unused O 2 .

2 C 2 H 6 + 7 O 2 → 4 CO 2 + 6 H 2 O In the balance equation given, the number of moles of oxygen is 3 . 5 times the number of moles of ethane. Hence, the volume of oxygen will also be thrice the amount of ethane. To burn 300 cc of ethane, the amount of oxygen needed = 3 . 5 × 300 = 1050 cc The amount of oxygen burnt with 300 cc of ethane in the given question = 1250 cc Hence, the amount of unused oxygen 1250 - 1050 cc = 200 cc of oxygen.

If 150 cc of gas A contains X molecules, how many molecules of gas B will be present in 75 cc of B ? The gases A and B are under the same conditions of temperature and pressure. Name the law on which the problem is based.

According to Avogadro’s law, equal volumes of all gases under similar conditions of temperature and pressure contain an equal number of molecules. So 150 cc of B will also contain X molecules. Hence, the number of molecules present in 75 cc of B will contain X 2 molecules.

Aluminium carbide reacts with water according to the following equation: Al 4 C 3 s + 12 H 2 O l → 4 Al OH 3 aq + 3 CH 4 g What volume of methane at s.t.p. is obtained from 12 g of aluminium carbide? [Relative molecular weight of Al 4 C 3 = 144 , Al OH 3 = 78 ]

Al 4 C 3 s + 12 H 2 O l → 4 Al OH 3 aq + 3 CH 4 g 144 g of Al 4 C 3 gives 3 × 12 + 4 × 1 = 48 g of methane. Therefore, 12 g of Al 4 C 3 gives = 12 × 48 144 = 4 g of CH 4 We know that 1 mole of any gas at STP occupies a volume of 22 . 4 L Therefore, the volume occupied by 4 g of CH 4 = 4 16 × 22 . 4 = 5 . 6 L

Aluminium carbide reacts with water according to the following equation: Al 4 C 3 s + 12 H 2 O l → 4 Al OH 3 aq + 3 CH 4 g . What mass of aluminium hydroxide is formed from 12 g of aluminium carbide?

Al 4 C 3 s + 12 H 2 O l → 4 Al OH 3 aq + 3 CH 4 g Atomic masses of the elements are: Al = 27 ; C = 12 ; O = 16 ; H = 1 From the above reaction, molecular mass of Al 4 C 3 = 144 g / mol Molecular mass of Al OH 3 = 78 g / mol Amount of aluminium hydroxide formed from 144 g of aluminium carbide = 4 27 × 1 + ( 12 + 1 ) × 3 = 312 g So, the amount of aluminium hydroxide formed from 12 g of aluminium carbide = 4 × 78 × 12 144 = 26 g

Copper sulphate solution reacts with sodium hydroxide solution to form a precipitate of copper hydroxide according to the equation: 2 NaOH + CuSO 4 → Na 2 SO 4 + Cu ( OH ) 2 ↓ i) What mass of copper hydroxide is precipitated by using 200 gm of sodium hydroxide? [ H = 1 , O = 16 , Na = 23 , S = 32 , Cu = 64 ] (ii) What is the colour of the precipitate formed?

(i) 2 NaOH + CuSO 4 → Na 2 SO 4 + Cu ( OH ) 2 ↓ Molecular mass of sodium hydroxide = 2 ( 23 + 16 + 1 ) = 80 g Molecular mass of copper hydroxide = 64 + 32 + 2 = 98 g 98 g of Cu ( OH ) 2 is precipitated by using 80 g of NaOH . x g of Cu ( OH ) 2 is precipitated by using 200 g of NaOH . x = 98 80 × 200 g = 245 g of Cu ( OH ) 2 (ii) A light blue precipitate of copper hydroxide will be formed.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 5 - Mole Concept and Stoichiometry>ICSE-2020>Q 11

HARD

10th ICSE

IMPORTANT

Earn 100

Calculate the number of molecules in $4.4 g$ of ${CO}_{2}$ .

[Atomic mass of $C = 12, O = 16$ ]

Important Questions on Mole Concept and Stoichiometry

HARD

10th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 5 - Mole Concept and Stoichiometry>ICSE-2020>Q 12

Calculate the amount of each reactant required to produce $750 ml$ of carbon dioxide, when two volumes of carbon monoxide combine with one volume of oxygen to produce two volumes of carbon dioxide.

$2 CO + O_{2} \to 2 {CO}_{2}$

State the law associated with this question.