Stoichiometry

Chemistry · 11th Class (Intermediate Part 1) · BISE Lahore

70 questions

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20 questions · Periodic Table and Periodic Properties × 1, Atomic Structure × 1, Chemical Bonding × 2, Stoichiometry × 1, States and Phases of Matter × 1, Chemical Energetics × 2, Reaction Kinetics × 1, Chemical Equilibrium × 1, Acid-Base Chemistry × 1, Electrochemistry × 2, Hydrocarbons × 2, Nitrogen and Sulfur × 1, Halogens × 1, Atmosphere × 1, Basic Separation Techniques × 1, Lab Safety and Practical Skills × 1

Stoichiometry Test

10 questions · Stoichiometry × 10

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Showing 10 of 70 questions from this chapter. The full bank is in the chapter test below.

1
The branch of chemistry that studies the quantitative relationship between the amounts of reactants and products in a balanced chemical equation is called:
- A.Stoichiometry (correct answer)
- B.Chemical kinetics
- C.Thermochemistry
- D.Electrochemistry
Why: Stoichiometry is derived from the Greek stoicheion (element) and metron (measure) and deals with the mass and mole relationships expressed by a balanced equation.
2
The Greek word 'stoicheion', from which the term stoichiometry is derived, means:
- A.measure
- B.element (correct answer)
- C.mixture
- D.reaction
Why: 'Stoicheion' means element while 'metron' means measure, so stoichiometry literally means the measurement of elements taking part in a chemical change.
3
One mole of a substance contains as many elementary entities as there are atoms in:
- A.1.0 g of hydrogen gas
- B.0.012 g of carbon-12
- C.0.012 kg of carbon-12 (correct answer)
- D.16.0 g of oxygen-16
Why: The mole is defined as the amount of substance containing the same number of entities as there are atoms in exactly 0.012 kg (12 g) of carbon-12.
4
The number of elementary entities present in one mole of any substance is:
- A. $6.02 \times 10^{22}$
- B. $6.02 \times 10^{24}$
- C. $6.02 \times 10^{-23}$
- D. $6.02 \times 10^{23}$ (correct answer)
Why: Avogadro's number $N_A$ equals $6.02 \times 10^{23}$ entities per mole, a rounded value of the exact constant $6.02214179 \times 10^{23}$ .
5
The molar mass of a substance is expressed in the unit:
- A.g
- B.g/mol (correct answer)
- C. $\pu{mol/dm3}$
- D.mol
Why: Molar mass is the mass of one mole of a substance, so its unit is grams per mole; $\pu{mol/dm3}$ is the unit of molar concentration instead.
6
What is the molar mass of $\ce{CCl4}$ ? (Ar: C = 12.0, Cl = 35.5)
- A.154.0 g/mol (correct answer)
- B.47.5 g/mol
- C.142.0 g/mol
- D.119.5 g/mol
Why: M = 12.0 + (4 x 35.5) = 12.0 + 142.0 = 154.0 g/mol; 142.0 g/mol is the mass of the chlorine atoms alone.
7
Calculate the number of moles present in 20 g of NaOH. (Molar mass of NaOH = 40 g/mol)
- A.0.25 mol
- B.2.0 mol
- C.20 mol
- D.0.50 mol (correct answer)
Why: n = m/M = 20 g / 40 $\pu{g mol-1}$ = 0.50 mol.
8
The mass of one mole of carbon dioxide is:
- A.28.0 g
- B.12.0 g
- C.44.0 g (correct answer)
- D.22.0 g
Why: $\ce{M(CO2)}$ = 12.0 + (2 x 16.0) = 44.0 g/mol, so one mole of $\ce{CO2}$ weighs 44.0 g; 28.0 g is the molar mass of CO.
9
Calculate the mass of 0.5 moles of HCl. (Molar mass of HCl = 36.5 g/mol)
- A.18.25 g (correct answer)
- B.36.5 g
- C.73.0 g
- D.7.3 g
Why: Mass = number of moles x molar mass = 0.5 mol x 36.5 $\pu{g mol-1}$ = 18.25 g.
10
The molar mass of $\ce{MgSO4}$ is 120 g/mol. What is the mass of $1 \times 10^{-3}$ mol of $\ce{MgSO4}$ ?
- A.12.0 g
- B.1.20 g
- C.0.12 g (correct answer)
- D.0.012 g
Why: Mass = n x M = $1 \times 10^{-3}$ mol x 120 $\pu{g mol-1}$ = 0.12 g.

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