Moles and Molar Calculations in Chemical Reactions
This is part of Year 11 HSC Chemistry course under the topic of Mole Concept.
HSC Chemistry Syllabus
- Explore the concept of the mole and relate this to Avogadro’s constant to describe, calculate and manipulate masses, chemical amounts and numbers of particles in: (ACSCH008, ACSCH039)
Moles and Molar Calculations
In this video, we will discuss calculation problems involving moles, molar mass and various chemical equations.
Practice Question 1 (Explanations in video)
2.00 g of solid sodium hydroxide is added to a solution of hydrochloric acid.
How many moles of hydrochloric acid will react with the solid sodium hydroxide?
Answer:
$$NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)$$
$$n(NaOH) = \frac{m}{MM} = \frac{2.00}{22.99+16.00+1.008} = 0.05 \text{ moles}$$
$$n(NaOH) = n(HCl) = 1:1$$
$$n(HCl) = 0.0500 \text{ moles (3 s.f)}$$
Practice Question 2 (Explanations in video)
The reaction between ethane gas (`C_2H_6`) and oxygen gas produces carbon dioxide and water.
How many moles of oxygen gas are required for a complete reaction with 200.0 g of ethane?
Answer:
$$C_2H_6(g) + \frac{7}{2} O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$$
or
$$2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(l)$$
$$nC_2H_6 = \frac{200.0}{12.01 \times 2 + 1.008 \times 6} = 6.65 \text{ moles}$$
$$n(O_2) = \frac{7}{2} n(C_2H_6) = 6.65 \times \frac{7}{2} = 23.28 \text{ moles (4 s.f)}$$
Practice Question 3 (Explanations in video)
When water is added to calcium, calcium hydroxide and hydrogen gas are produced.
How many molecules of water will react with 4.76 g of calcium?
Answer:
$$2H_2O(l) + Ca(s) \rightarrow Ca(OH)_2(aq) + H_2(g)$$
$$n(Ca) = \frac{4.76}{40.08} = 0.119 \text{ moles}$$
$$n(H_2O) = 2 \times n(Ca) = 0.119 \times 2 = 0.238 \text{ moles}$$
$$\text{Molecules = moles } \times \text{Avogadro's number}$$
$$\text{Number of water molecules} = 0.238 \times 6.022 \times 10^{23} = 1.43 \times 10^{23} \text{ molecules}$$
Practice Question 4 (Explanations in video)
50.0 g of water is added to 43.0 g of sodium to produce sodium hydroxide and hydrogen gas.
How many moles of hydrogen gas are produced from this reaction?
Answer:
$$2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)$$
$$n(H_2O) = \frac{50.0}{1.008 \times 2 + 16} = 2.78 \text{ moles}$$
$$n(Na) = \frac{43}{22.99} = 1.87 \text{ moles}$$
$$n(H_2O) : n(Na) = 1:1$$
$$n(Na)< n(H_2O)$$
$$ \therefore \text{ Na is the limiting reagent}$$
$$n(H_2) = \frac{1}{2} n(Na) = 1.87 \div 2 = 0.935 \text{ moles (3 s.f)}$$
Practice Question 5 (Explanations in video)
2.00 mol of nitric acid is added to a 50.0 g pure sample of magnesium carbonate.
How many molecules of carbon dioxide are produced from this reaction?
Answer:
$$2HNO_3(aq) + MgCO_3(s) \rightarrow Mg(NO_3)_2 + CO_2(g) + H_2O(l)$$
$$n(HNO_3) = 2.00 \text{ moles (Given in question)}$$
$$n(MgCO_3) = \frac{50.0}{24.31+12.01+16 \times 3} = 0.593 \text{ moles}$$
$$n(HNO_3)_{\text{required}} = 0.593 \times 2 = 1.186 \text{ moles}$$
Amount of nitric acid required is less than what's available.
$$\therefore MgCO_3 \text{ is the limiting reagent}$$
$$n(CO_2) = n(MgCO_3) = 0.593 \text{ moles (1:1 reaction ratio)}$$
$$\text{Molecules of } CO_2 = 0.593 \times 6.022 \times 10^{23} = 3.57 \times 10^{23} \text{ molecules (3 s.f)}$$
Practice Question 6 (Explanations in video)
Calcium carbonate can be found in limestone. Excess hydrochloric acid is added to a 10.0 g sample of limestone. It was determined that 0.070 moles of carbon dioxide were formed.
What is the percentage by mass of the calcium carbonate in the sample of limestone?
Answer:
$$CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)$$
$$n(CO_2) = 0.070 \text{ moles}$$
$$n(CaCO_3) = n(CO_2) = 0.070 \text{ moles (1:1 reaction ratio)}$$
$$m(CaCO_3) = 0.0070 \times (40.08 + 12.01 + 16 \times 3) = 7.0 \text{ g}$$
$$ \text{Percentage by mass} = \frac{7.0}{10.0} \times 100 = 70 \text{% (2 s.f.)}$$
Practice Question 7 (Explanations in video)
A particular element X can be found in a compound with chlorine in the form of `XCl_2` and `XCl_4`. Treatment of 10.00 g of `XCl_2` with excess chlorine gas formed 12.55 g of `XCl_4`.
Identify element X. Support your answer with calculations.
Answer:
$$XCl_2(?) + Cl_2(g) \rightarrow XCl_4(?)$$
$$m(Cl_2) = m(XCl_4) – m(XCl_2) = 12.55\text{ g} – 10.00\text{ g} = 2.55\text{ g}$$
$$n(Cl_2) = \frac{2.55}{35.45 \times 2} = 0.0360 \text{ moles}$$
$$n(Cl_2) = n(XCl_2) = 1:1$$
$$\therefore n(XCl_2) = 0.0360 \text{ moles}$$
$$MM(XCl_2) = \frac{10.00 \text{ g}}{0.0360 \text{ moles}} = 278.04 \text{ g } mol^{–1}$$
$$MM(X) = 278.04 – 35.45 \times 2 = 207.1 \text{ g}$$
$$\therefore \text{ element X is Pb (lead) as it has the closest molar mass to 207.1 g}$$