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Secondary 4 Combined Science Chemistry Stoichiometry Moles Quiz

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Secondary 4 Combined Science Chemistry Quiz - Stoichiometry Moles

Name: ____________________
Class: ____________________
Date: ____________________
Score: ________ / 45

Duration: 60 Minutes
Total Marks: 45 Marks

Instructions:

Answer all questions.
Show all working for calculation questions.
Give your answers to 3 significant figures where applicable.
Relative Atomic Masses ( $A_r$ ): H=1, C=12, N=14, O=16, Na=23, Mg=24, S=32, Cl=35.5, K=39, Ca=40.

Section A: Fundamental Concepts (Questions 1-5)

Define the term 'mole' in chemistry. [1]
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Calculate the relative molecular mass ( $M_r$ ) of glucose, $\text{C}_6\text{H}_{12}\text{O}_6$ . [1]
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Calculate the mass of 0.25 mol of sodium carbonate, $\text{Na}_2\text{CO}_3$ . [2]

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How many moles of atoms are present in 0.1 mol of $\text{H}_2\text{SO}_4$ ? [1]
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Determine the relative molecular mass of urea, $\text{CO}(\text{NH}_2)_2$ . [1]
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Section B: Gas Volumes and Concentrations (Questions 6-12)

Calculate the volume occupied by 0.08 mol of nitrogen gas at room temperature and pressure (RTP). [2]
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A sample of helium gas occupies $1.2\text{ dm}^3$ at RTP. Calculate the number of moles of helium present. [2]
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Calculate the concentration in $\text{mol/dm}^3$ of a solution prepared by dissolving 4.0g of $\text{NaOH}$ in $250\text{ cm}^3$ of distilled water. [3]

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What mass of potassium chloride ( $\text{KCl}$ ) is required to prepare $100\text{ cm}^3$ of a $0.5\text{ mol/dm}^3$ solution? [3]

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A gas occupies $3.6\text{ dm}^3$ at RTP and has a molar mass of $44\text{ g/mol}$ . Identify the gas and calculate its mass in the sample. [3]
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Calculate the number of moles of solute in $50\text{ cm}^3$ of a $2.0\text{ mol/dm}^3$ solution of $\text{HCl}$ . [2]
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Which has more particles: 2.0g of Hydrogen gas ( $\text{H}_2$ ) or 2.0g of Oxygen gas ( $\text{O}_2$ )? Explain your answer. [3]
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Section C: Stoichiometry and Chemical Equations (Questions 13-20)

Balance the following equation: $\text{Al}(\text{s}) + \text{O}_2(\text{g}) \rightarrow \text{Al}_2\text{O}_3(\text{s})$ . [1]
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For the reaction: $\text{Mg}(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MgCl}_2(\text{aq}) + \text{H}_2(\text{g})$ , calculate the mass of magnesium that reacts completely with $0.2\text{ mol}$ of $\text{HCl}$ . [3]

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$2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O}(\text{l})$ . Calculate the volume of oxygen gas required at RTP to react completely with $10\text{ dm}^3$ of hydrogen gas. [2]
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Calcium carbonate decomposes upon heating: $\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})$ . Calculate the mass of $\text{CaCO}_3$ needed to produce $1.2\text{ dm}^3$ of $\text{CO}_2$ at RTP. [4]

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Zinc reacts with sulfuric acid: $\text{Zn}(\text{s}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{ZnSO}_4(\text{aq}) + \text{H}_2(\text{g})$ . If $6.5\text{g}$ of Zinc is used, calculate the volume of $\text{H}_2$ gas produced at RTP. ( $A_r \text{ Zn} = 65$ ). [3]

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A $0.1\text{ mol/dm}^3$ solution of $\text{NaOH}$ is reacted with $\text{HCl}$ : $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ . Calculate the volume of $\text{NaOH}$ solution needed to neutralize $25\text{ cm}^3$ of $0.2\text{ mol/dm}^3 \text{HCl}$ . [3]

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In the reaction $2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3$ , if $0.3\text{ mol}$ of $\text{Al}$ reacts with $0.4\text{ mol}$ of $\text{Cl}_2$ , identify the limiting reactant. [3]

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Calculate the mass of $\text{AlCl}_3$ formed in Question 19. [4]

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Answers

Answer Key - Stoichiometry Moles Quiz

Definition: The amount of substance that contains as many elementary particles (atoms, molecules, ions) as there are atoms in exactly 12g of carbon-12. [1]
$M_r$ of Glucose: $(6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180$ . [1]
Mass of $\text{Na}_2\text{CO}_3$ :
- $M_r = (2 \times 23) + 12 + (3 \times 16) = 46 + 12 + 48 = 106$
- $\text{Mass} = \text{moles} \times M_r = 0.25 \times 106 = 26.5\text{g}$ . [2]
Moles of atoms:
- 1 molecule of $\text{H}_2\text{SO}_4$ contains $2+1+4 = 7$ atoms.
- Total moles of atoms $= 0.1 \times 7 = 0.7\text{ mol}$ . [1]
$M_r$ of Urea: $12 + 16 + 2(14 + 2 \times 1) = 28 + 2(16) = 28 + 32 = 60$ . [1]
Volume of $\text{N}_2$ :
- $\text{Volume} = \text{moles} \times 24\text{ dm}^3/\text{mol} = 0.08 \times 24 = 1.92\text{ dm}^3$ . [2]
Moles of He:
- $\text{Moles} = \text{Volume} / 24 = 1.2 / 24 = 0.05\text{ mol}$ . [2]
Concentration of $\text{NaOH}$ :
- $M_r(\text{NaOH}) = 23 + 16 + 1 = 40$
- $\text{Moles} = 4.0 / 40 = 0.1\text{ mol}$
- $\text{Volume} = 250 / 1000 = 0.25\text{ dm}^3$
- $\text{Conc} = 0.1 / 0.25 = 0.4\text{ mol/dm}^3$ . [3]
Mass of $\text{KCl}$ :
- $M_r(\text{KCl}) = 39 + 35.5 = 74.5$
- $\text{Moles} = \text{Conc} \times \text{Vol} = 0.5 \times (100/1000) = 0.05\text{ mol}$
- $\text{Mass} = 0.05 \times 74.5 = 3.725\text{g} \approx 3.73\text{g}$ . [3]
Gas Identification:
- $\text{Moles} = 3.6 / 24 = 0.15\text{ mol}$
- $\text{Mass} = 0.15 \times 44 = 6.6\text{g}$
- Gas is Carbon Dioxide ( $\text{CO}_2$ ). [3]
Moles of $\text{HCl}$ :
- $\text{Moles} = 2.0 \times (50/1000) = 0.1\text{ mol}$ . [2]
Particle Comparison:
- $\text{Moles of } \text{H}_2 = 2.0 / 2 = 1.0\text{ mol}$
- $\text{Moles of } \text{O}_2 = 2.0 / 32 = 0.0625\text{ mol}$
- $\text{H}_2$ has more particles because it has a lower molar mass, resulting in more moles for the same mass. [3]
Balanced Equation: $4\text{Al}(\text{s}) + 3\text{O}_2(\text{g}) \rightarrow 2\text{Al}_2\text{O}_3(\text{s})$ . [1]
Mass of Mg:
- Ratio $\text{Mg} : \text{HCl} = 1 : 2$
- $\text{Moles of Mg} = 0.2 / 2 = 0.1\text{ mol}$
- $\text{Mass} = 0.1 \times 24 = 2.4\text{g}$ . [3]
Volume of $\text{O}_2$ :
- Ratio $\text{H}_2 : \text{O}_2 = 2 : 1$
- $\text{Volume of } \text{O}_2 = 10\text{ dm}^3 / 2 = 5.0\text{ dm}^3$ . [2]
Mass of $\text{CaCO}_3$ :
- $\text{Moles of } \text{CO}_2 = 1.2 / 24 = 0.05\text{ mol}$
- Ratio $\text{CaCO}_3 : \text{CO}_2 = 1 : 1 \rightarrow 0.05\text{ mol } \text{CaCO}_3$
- $M_r(\text{CaCO}_3) = 40 + 12 + (3 \times 16) = 100$
- $\text{Mass} = 0.05 \times 100 = 5.0\text{g}$ . [4]
Volume of $\text{H}_2$ :
- $\text{Moles of Zn} = 6.5 / 65 = 0.1\text{ mol}$
- Ratio $\text{Zn} : \text{H}_2 = 1 : 1 \rightarrow 0.1\text{ mol } \text{H}_2$
- $\text{Volume} = 0.1 \times 24 = 2.4\text{ dm}^3$ . [3]
Volume of $\text{NaOH}$ :
- $\text{Moles of } \text{HCl} = 0.2 \times (25/1000) = 0.005\text{ mol}$
- Ratio $\text{NaOH} : \text{HCl} = 1 : 1 \rightarrow 0.005\text{ mol } \text{NaOH}$
- $\text{Volume} = \text{moles} / \text{conc} = 0.005 / 0.1 = 0.05\text{ dm}^3 = 50\text{ cm}^3$ . [3]
Limiting Reactant:
- Required $\text{Cl}_2$ for $0.3\text{ mol Al} = 0.3 \times (3/2) = 0.45\text{ mol}$
- Available $\text{Cl}_2 = 0.4\text{ mol}$
- Since $0.4 < 0.45$ , $\text{Cl}_2$ is the limiting reactant. [3]
Mass of $\text{AlCl}_3$ :
- Use limiting reactant $\text{Cl}_2$ : Ratio $\text{Cl}_2 : \text{AlCl}_3 = 3 : 2$
- $\text{Moles of } \text{AlCl}_3 = 0.4 \times (2/3) = 0.267\text{ mol}$
- $M_r(\text{AlCl}_3) = 27 + (3 \times 35.5) = 133.5$
- $\text{Mass} = 0.267 \times 133.5 = 35.6\text{g}$ . [4]