Complete the following blanks in the equation as indicated CaH 2 ( s ) + 2 H 2 O ( aq ) → Ca ( OH ) 2 ( s ) + 2 H 2 ( g ) . Molecules: 6 . 023 × 10 23 + _____ → _____ + _____.

In the field of chemistry, a mole is defined as the amount of a substance that contains exactly 6 . 023 × 10 23 ‘elementary entities’ of the given substance. From the given equation, CaH 2 ( s ) + 2 H 2 O ( aq ) → Ca ( OH ) 2 ( s ) + 2 H 2 ( g ) , the blanks can be filled as follows, Molecules: 6 . 023 × 10 23 + 2 × 6 . 023 × 10 23 → 6 . 023 × 10 23 + 2 × 6 . 023 × 10 23 .

The reaction between 15 g of marble and nitric acid is given by the following equation, CaCO 3 + 2 HNO 3 → Ca ( NO 3 ) 2 + H 2 O + CO 2 . Calculate the mass of anhydrous calcium nitrate formed.

The given equation, CaCO 3 + 2 HNO 3 → Ca ( NO 3 ) 2 + H 2 O + CO 2 , 15 g of marble and nitric acid are reacted. One mole of calcium carbonate reacts with two moles of nitric acid form one mole of anhydrous calcium nitrate. It is the decarboxylation reaction. Therefore, number of moles of calcium carbonate reacted, = Given mass Molar mass of CaCO 3 = 15 40 + 12 + 3 × 16 = 15 100 = 0 . 15 mol . So, 0 . 15 mol of calcium carbonate reacted with nitric acid produce 0 . 15 mol of anhydrous calcium nitrate. i.e., 0 . 15 mol = x g 40 + 2 14 + 3 16 . 0 . 15 mol = x g 164 g mol - 1 . x g = 0 . 15 × 164 g . Thus, 24 . 6 g of calcium nitrate can be formed.

The reaction between 15 g of marble and nitric acid is given by the following equation, CaCO 3 + 2 HNO 3 → Ca ( NO 3 ) 2 + H 2 O + CO 2 . Calculate the volume of carbon dioxide evolved at STP.

The given equation, CaCO 3 + 2 HNO 3 → Ca ( NO 3 ) 2 + H 2 O + CO 2 , 15 g of marble and nitric acid are reacted. One mole of calcium carbonate reacts with two moles of nitric acid evolves one mole of carbon dioxide. Therefore, number of moles of calcium carbonate reacted, = Given mass Molar mass of CaCO 3 = 15 40 + 12 + 3 × 16 = 15 100 = 0 . 15 mol . So, 0 . 15 mol of calcium carbonate reacted with nitric acid produce 0 . 15 mol of carbon dioxide. i.e., 0 . 15 mol = x L 22 . 4 L . x L = 0 . 15 × 22 . 4 L = 3 . 36 L . Thus, 3 . 36 L of carbon dioxide is evolved.

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

E M B I B E

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5E>Q 21

EASY

10th ICSE

IMPORTANT

Earn 100

A compound with empirical formula $AB$ has vapour density three times its empirical formula weight. Find the molecular formula.

Important Questions on Mole Concept and Stoichiometry

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5E>Q 21

10.47 g

of a compound contained

6.25 g

of metal

A

and rest non-metal

B

. Calculate the empirical formula of the compound (At. wt of

A = 207; B = 35.5

).

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5E>Q 22

A hydride of nitrogen contains

87.5

per cent by mass of nitrogen. Determine the empirical formula of this compound.

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5E>Q 23

A compound has

O = 61.32 %; S = 11.15 %; H = 4.88 %

and

Zn = 22.65 %

. The relative molecular mass of the compound is

287 a . m . u

. Find the molecular formula of the compound, assuming that all the hydrogen is present as water of crystallisation.

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5D>Q 1

Complete the following blanks in the equation as indicated ${CaH}_{2} (s) + 2 H_{2} O (aq) \to Ca {(OH)}_{2} (s) + 2 H_{2} (g)$ .

Moles: $1 mole +$ _____ $\to$ _____ $+$ _____.

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5D>Q 1

Complete the following blanks in the equation based on ${CaH}_{2} (s) + 2 H_{2} O (aq) \to Ca {(OH)}_{2} (s) + 2 H_{2} (g)$ .

Grams: $42 g +$ _____ $\to$ _____ $+$ _____.

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5D>Q 1

Complete the following blanks in the equation as indicated ${CaH}_{2} (s) + 2 H_{2} O (aq) \to Ca {(OH)}_{2} (s) + 2 H_{2} (g)$ .

Molecules: $6.023 \times 10^{23} +$ _____ $\to$ _____ $+$ _____.

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5D>Q 2

The reaction between $15 g$ of marble and nitric acid is given by the following equation, ${CaCO}_{3} + 2 {HNO}_{3} \to Ca {({NO}_{3})}_{2} + H_{2} O + {CO}_{2}$ . Calculate the mass of anhydrous calcium nitrate formed.

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5D>Q 2

The reaction between $15 g$ of marble and nitric acid is given by the following equation, ${CaCO}_{3} + 2 {HNO}_{3} \to Ca {({NO}_{3})}_{2} + H_{2} O + {CO}_{2}$ . Calculate the volume of carbon dioxide evolved at STP.