560 mL of carbon monoxide is mixed with 500 mL of oxygen and ignited. The chemical equation for the reaction is as follows: 2 CO + O 2 → 2 CO 2 Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.

Consider the given reaction, 2 CO + O 2 → 2 CO 2 . According to the given balanced equation, 2 × 22400 mL of carbon monoxide reacts with 22400 mL of oxygen. Therefore, 560 mL of carbon monoxide reacts with 560 × 22400 2 × 22400 = 280 mL of oxygen. Again, 2 × 22400 mL of carbon monoxide produces 2 × 22400 mL of carbon dioxide. Hence, 560 mL of carbon monoxide produces 560 × 2 × 22400 2 × 22400 = 560 mL of carbon dioxide.

A gas cylinder contains 24 × 10 24 molecules of nitrogen gas. If Avogadro’s number is 6 × 10 23 and the relative atomic mass of nitrogen is 14 , calculate the mass of nitrogen gas in the cylinder.

We know that, the mass of 6 × 10 23 molecules of nitrogen = 28 g Therefore, the mass of 24 × 10 24 molecules of nitrogen = 24 × 10 24 × 28 6 × 10 23 = 1120 g .

A gas cylinder contains 24 × 10 24 molecules of nitrogen gas. If Avogadro’s number is 6 × 10 23 and the relative atomic mass of nitrogen is 14 , calculate the volume of nitrogen at STP in dm 3 .

We know that, 6 × 10 23 molecules of nitrogen occupies the volume of 22 . 4 dm 3 at STP . Therefore, 24 × 10 24 molecules of nitrogen occupies the volume of = 24 × 10 24 × 22 . 4 6 × 10 23 = 896 dm 3 at STP .

A certain gas ‘ X ’ occupies a volume of 100 cm 3 at STP and weighs 0 . 5 g . Find its relative molecular mass.

Given, 0 . 5 g of a gas X occupies 100 cm 3 at STP . Therefore, the mass of the gas which occupies 22400 cm 3 at STP = Gram molecular mass = 22400 × 0 . 5 100 = 112 g Therefore, relative molecular mass of the gas 112 g .

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.

Calculate the number of moles in 12 g of oxygen gas. [ O = 16 ]

Molecular mass = 16 × 2 = 32 g 32 g of oxygen gas → 1 mole 1 g of oxygen gas → 1 32 mole Given that 12 g of oxygen gas Number of moles = 12 32 = 0 . 375 mole

MEDIUM

10th ICSE

IMPORTANT

Earn 100

Calculate the percentage of sodium in sodium aluminium fluoride ( ${Na}_{3} {AlF}_{6}$ ). Correct to the nearest whole number. ( $F = 19, Na = 23, Al = 27$ )

Important Questions on Mole Concept and Stoichiometry

MEDIUM

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>MISCELLANEOUS EXERCISE>Q 46

$560 mL$ of carbon monoxide is mixed with $500 mL$ of oxygen and ignited. The chemical equation for the reaction is as follows:

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

Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.

MEDIUM

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>MISCELLANEOUS EXERCISE>Q 47

A gas cylinder of the capacity of

20 {dm}^{3}

is filled with gas

X

the mass of which is

10 g

. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogen is

2 g

, hence the relative molecular mass of the gas is:

MEDIUM

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>MISCELLANEOUS EXERCISE>Q 47

Calcium carbide is used for the artificial ripening of fruits. Actually, the fruit ripens because of the heat evolved while calcium carbide reacts with moisture. During this reaction calcium hydroxide and acetylene gas is formed. If $200 {cm}^{3}$ of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below.

$2 C_{2} H_{2} (g) + 5 O_{2} (g) \overset{}{\to} 4 {CO}_{2} (g) + 2 H_{2} O (g)$

MEDIUM

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>MISCELLANEOUS EXERCISE>Q 47

A gaseous compound of nitrogen and hydrogen contains

12.5 %

hydrogen by mass. Find the molecular formula of the compound if its relative molecular mass is

37

MEDIUM

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>MISCELLANEOUS EXERCISE>Q 47

A gas cylinder contains $24 \times 10^{24}$ molecules of nitrogen gas. If Avogadro’s number is $6 \times 10^{23}$ and the relative atomic mass of nitrogen is $14$ , calculate the mass of nitrogen gas in the cylinder.

MEDIUM

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>MISCELLANEOUS EXERCISE>Q 47

A gas cylinder contains $24 \times 10^{24}$ molecules of nitrogen gas. If Avogadro’s number is $6 \times 10^{23}$ and the relative atomic mass of nitrogen is $14$ , calculate the volume of nitrogen at $STP$ in ${dm}^{3}$ .

MEDIUM

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>MISCELLANEOUS EXERCISE>Q 47

Commercial sodium hydroxide weighing $30 g$ has some sodium chloride in it. The mixture on dissolving in water and subsequent treatment with excess of silver nitrate solution formed a precipitate weighing $14.3 g$ . What is the percentage of sodium chloride in the commercial sample of sodium hydroxide? The equation for the reaction is $NaCl + {AgNO}_{3} \overset{}{\to} AgCl + {NaNO}_{3}$ .

(Relative molecular mass of $NaCl = 58, AgCl = 143$ )

MEDIUM

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>MISCELLANEOUS EXERCISE>Q 47

A certain gas ‘

X

’ occupies a volume of

100 {cm}^{3}

STP

and weighs

0.5 g

. Find its relative molecular mass.

Calculate the percentage of sodium in sodium aluminium fluoride (Na3AlF6). Correct to the nearest whole number. (F=19, Na=23, Al=27)

Important Questions on Mole Concept and Stoichiometry

Learn and Prepare for any exam you want !

Calculate the percentage of sodium in sodium aluminium fluoride ( ${Na}_{3} {AlF}_{6}$ ). Correct to the nearest whole number. ( $F = 19, Na = 23, Al = 27$ )