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Iron forms a compound that contains iron (36.8%), sulfur (31.6%), and oxygen (31.6%). The compound boils at 120°C. For one mole of this compound, the density of its vapor at 150°C and 250 kPa is 42.0 g/L.
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a. \(\ce{Fe2S3O6}\)
b. \(590.94\ \text{g mol}^{-1}\).
a. Divide each compound’s percentage by their molar masses:
\(\ce{Fe}: \dfrac{36.8%}{55.85} = 0.65 \ \Rightarrow \ \dfrac{0.659}{0.659}=1\)
\(\ce{S}: \dfrac{31.6%}{32.07} = 0.985 \ \Rightarrow \ \dfrac{0.985}{0.659}=\dfrac{3}{2}\)
\(\ce{O}: \dfrac{31.6%}{16.00} = 1.975 \ \Rightarrow \ \dfrac{1.975}{0.659}=3\)
→ Due to the fraction, each number must be doubled so there are only whole numbers.
→ Thus the empirical formula for the compound is \(\ce{Fe2S3O6}\).
b. Using the Ideal Gas Law to find the volume of the vapour:
\(V=\dfrac{nRT}{P}=\dfrac{1 \times 8.314 \times (150+273)}{250}=14.07\ \text{L}\)
→ Molar mass of the vapour \(= 42 \times 14.07 = 590.94\ \text{g mol}^{-1}\).
Aluminium carbonate reacts with nitric acid to produce aluminium nitrate, carbon dioxide, and water.
What volume of carbon dioxide will be produced if 15.0 g of aluminium carbonate is reacted and the gas is collected at 25°C and 100 kPa? (5 marks)
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\(4.76\ \text{L}\)
→ The chemical equation for the reaction:
\(\ce{Al2(CO3)3 + 6HNO3 -> 2Al(NO3)3 + 3CO2 + 3H2O}\)
→ \(\ce{n(Al2(CO3)3)}=\dfrac{m}{MM}=\dfrac{15.0}{2(26.98+3(12.01)+9(16)}=0.0641\ \text{mol}\)
→ The ratio of \(\ce{Al2(CO3)3:CO2 = 1:3}\)
→ \(\ce{n(CO2)}= 3 \times 0.0641 = 0.192\ \text{mol}\)
→ Using Avogadro’s Law:
\(V=n \times V_{\text{molar}} = 0.192 \times 24.79 = 4.76\ \text{L}\)
A gas at a temperature of \(9.0 \times 10^2\ \text{K}\) in a container with a volume of \(30.0\ \text{L}\) has a pressure of \(5.0 \times 10^2\ \text{kPa}\).
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a. \(2.0\ \text{mol}\)
b. The unknown gas is fluorine.
a. Using the Ideal Gas Law:
\(PV=nRT \Rightarrow n=\dfrac{PV}{RT}\)
\(n=\dfrac{500 \times 30}{8.314 \times 900}=2.0\ \text{mol (2 sig.fig)}\)
b. Mass of the gas \(=526.0-450.0=76\ \text{g}\)
Molar Mass of the gas \(=\dfrac{m}{n}=\dfrac{76}{2.0}=38\ \text{g mol}^{-1}\)
→ This is equal to the MM of fluorine gas \(\ce{(F2)}\).
→ The unknown gas is fluorine.
Use the equation below to answer the following two questions:
\(\ce{2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l)}\)
Part 1
What volume of carbon dioxide would be produced by the combustion of 3.0 L of ethane gas \(\ce{C2H6(g)}\) in excess oxygen? (Assume the temperature and pressure remain the same)
Part 2
Whose law is applied when determining the volume relationships in the reaction above?
Part 1: \(A\)
Part 2: \(C\)
Part 1
→ From the chemical equation, the mole ratio of ethane to carbon dioxide is \(1:2\).
→ According to Avogadro’s Law, at the same temperature and pressure, the volume of gases is directly proportional to the number of moles.
→ Thus, from the equation, 1 volume of ethane produces 2 volumes of carbon dioxide.
→ The volume of carbon dioxide produced \(=2 \times 3.0 = 6.0\ \text{L}\)
\(\Rightarrow A\)
\(\text{Part 2}\)
→ When determining the volume relationships between gases in a chemical reaction, Avogadro’s Law is used.
→ Avogadro’s Law states that equal volumes of gases, at the same temperature and pressure, contain the same number of moles of gas.
→ The volumes of gas react in simple whole-number ratios, as shown in the balanced equation.
\(\Rightarrow C\)
A piece of zinc weighing 3.20 grams is placed into a beaker containing 300.0 mL of 0.7500 mol/L hydrochloric acid.
\(\ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)}\)
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a. Zinc is the limiting reagent.
b. \(1.21\ \text{L}\)
a. \(\ce{n(Zn)} = \dfrac{m}{MM} = \dfrac{3.20}{65.38}=0.0489\ \text{mol}\)
\(\ce{n(HCl)} = c \times V = 0.75 \times 0.3 = 0.225\)
→ Based on the mole ratio, \(0.0489\ \text{mol}\) of Zinc would require \(0.0978\ \text{mol}\) of hydrochloric acid.
→ As there is excess hydrochloric acid, Zinc is the limiting reagent.
b. The Mole ratio of \(\ce{Zn:H2}\) is \(1:1\)
→ \(0.0489\ \text{mol}\) of \(\ce{H2(g)}\) is produced.
→ At \(25^{\circ}\text{C}\) and \(100\ \text{kPa}\), \(1\) mole of gas \(=24.79\ \text{L}\)
→ \(\ce{V(H2(g))} = 0.0489 \times 24.79 = 1.21\ \text{L}\)
Carbon dioxide is produced during the combustion of propane \(\ce{(C3H8)}\) in oxygen \(\ce{(O2)}\). The balanced chemical equation for this reaction is:
\(\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)}\)
If 44.0 grams of propane are completely combusted, calculate the volume of carbon dioxide produced at STP (100 kPa and 0\(^{\circ}\)C). (3 marks)
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\(\ce{V(CO2)}\ =68\ \text{L}\)
Calculate the number of moles of propane (\(\ce{C3H8}\)):
\(\ce{n(C3H8) = \dfrac{\text{m}}{\text{MM}} = \dfrac{44.0\ \text{g}}{44.094\ \text{g mol}^{-1}} = 0.998\ \text{mol}}\)
Use the stoichiometric ratio to find the moles of \(\ce{CO2}\) produced:
\(\ce{n(CO2) = 3 \times n(C3H8) = 3 \times 0.998 \, mol = 2.994 \, mol}\)
Calculate the volume of \(\ce{CO2}\) at STP:
\(\ce{V(CO2) = n \times 22.71 \, L \, mol^{-1} = 2.994 \, mol \times 22.71 \, L \, mol^{-1} = 68 \, L}\)
→ The volume of carbon dioxide produced is 68 litres.
A helium balloon is inflated to a volume of 5.65 L and a pressure of 10.2 atm at a temperature of 25°C.
Calculate the amount of helium, in moles, in the balloon. (2 marks)
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\(2.36\ \text{mol}\)
| \(\ce{n(He)}\) | \(= \dfrac{pV}{RT} \) | |
| \(= \dfrac{10.2 \times 101.3 \times 5.65}{8.31 \times 298} \) | ||
| \(= 2.36\ \text{mol} \) |
The emergency oxygen system in a passenger aircraft uses the decomposition of sodium chlorate to produce oxygen.
At 76.0 kPa and 292 K, each adult passenger needs about 1.60 L of oxygen per minute. The equation for the reaction is
\(\ce{2NaClO3(s)\rightarrow 2NaCl(s) + 3O2(g)}\)
\(\ce{MM(NaClO3) = 106.5 \text{g mol}^{–1}}\)
Calculate the mass of sodium chlorate required to provide the required volume of oxygen for each adult passenger per minute. (3 marks)
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\(\text{3.56 grams\)
\(\ce{2NaClO3(s)\rightarrow 2NaCl(s) + 3O2(g)}\)
\(\ce{n(O2)_{\text{req}} = \dfrac{pV}{RT} = \dfrac{76.0 \times 1.60}{8.31 \times 292} = 0.0501\ \text{mol} }\)
\(\ce{n(NaClO3) = \dfrac{2}{3} \times n(O2) = \dfrac{2}{3} \times 0.0501 = 0.0334\ \text{mol}}\)
\(\ce{m(NaClO3)_{\text{req}} = n \times MM = 0.0334 \times 106.6 = 3.56\ \text{g min}^{-1}}\)
A gas is produced when 10.0 g of zinc is placed in 0.50 L of 0.20 mol L\(^ {-1}\) nitric acid.
Calculate the volume of gas produced at 25°C and 100 kPa. Include a balanced chemical equation in your answer. (4 marks)
\(1.24\ \text{L}\)
\(\ce{2HNO3 + Zn \rightarrow Zn(NO3)2 + H2}\)
\(\ce{n(HNO3) = c \times V = 0.20 \times 0.5 = 0.10 moles}\)
\(\ce{n(Zn) = \dfrac{\text{m}}{\ce{MM}} = \dfrac{10}{65.38} = 0.153 moles}\)
→ \(\ce{HNO3}\) and \(\ce{Zn}\) react in a \(2:1\) ratio
→ \(\ce{HNO3}\) is the limiting reagent and 0.05 moles of gas \(\ce{H2}\) will be produced
\(\ce{V(H2) = 0.05 \times 24.79 = 1.24\ \text{L (2 d.p.)}}\)
Sodium azide is used in automobile airbags to provide a source of nitrogen gas for rapid inflation in an accident.The equation shows the production of nitrogen gas from sodium azide.
\( \ce{2NaN3}(s) \rightarrow \ce{2Na}(s) + \ce{3N2}(g)\)
What mass of sodium azide will produce 40L of \(\ce{N2}\) at 100 kPa and 0°C?
\(B\)
→ At 100 kPa and 0°C, constant is 22.71 \(\text{Lmol}^{-1}\)
\(n(\ce{N2})=\dfrac{40}{22.71}=1.76\ \text{mol}\)
\(n(\text{sodium azide})= 1.76 \times \dfrac{2}{3} =1.17\ \text{mol}\)
\(m(\text{sodium azide})=1.17 \times 65.02 =76\ \text{g}\)
\(\Rightarrow B\)
Sodium reacts with water to give hydrogen gas and sodium hydroxide solution.
What volume of gas would be produced from the reaction of 22.99 g of sodium at 25°C and 100 kPa ?
\(B\)
\(\ce{2Na(s) + 2H2O(l) \rightarrow H2(g) + 2NaOH(aq)}\)
\(n(\ce{Na(s)}) = \dfrac{22.99}{22.99}= 1\ \text{mol}\)
\(n(\ce{H2(g)}) = \dfrac{1}{2} \times 1= 0.5\ \text{mol}\)
The volume of \(\ce{H2(g)}= 0.5 \times 24.79 = 12.40\ \text{L}\)
\(\Rightarrow B\)
What is the density of ozone \(\ce{(O3(g))}\) at 25°C and 100 kPa ?
\(C\)
→ Molar Mass of ozone \(= 3 \times 16=48\ \text{gmol}^{-1}\).
→ Density at 25°C and 100 kPa is 24.79 \(\text{Lmol}^{-1}\)
→ Density in \(\text{gL}^{-1}= \dfrac{48}{24.79}=1.936\ \text{gL}^{-1}\)
\(\Rightarrow C\)
An unattended car is stationary with its engine running in a closed workshop. The workshop is 5.0 m × 5.0 m × 4.0 m and its volume is \(1.0\ ×\ 10^{5}\) L. The engine of the car is producing carbon monoxide in an incomplete combustion according to the following chemical equation:
\(\ce{C8H18(l)+\dfrac{17}{2}O2(g) \rightarrow 8CO(g) + 9H2O(l)}\)
Exposure to carbon monoxide at levels greater than 0.100 g L\(^{-1}\) of air can be dangerous to human health.
6.0 kg of octane was combusted by the car in this workshop.
Using the equation provided, determine if the level of carbon monoxide produced in the workshop would be dangerous to human health. Support your answer with relevant calculations. (4 marks)
\(\ce{\text{Volume of garage} = 5 \times 5 \times 4 = 100 \text{m}^2}\)
\(\ce{100 \text{m}^2} = 100\ 000\ \text{L}\)
\[\ce{\text{n(octane)} = \frac{n}{M} = \frac{6000}{114.224} = 52.53 \text{moles}} \]
\(\ce{Molar ratio of Octane : CO = 1:8}\)
\(\ce{n(CO) = 8 \times 52.53 = 420.23}\)
\(\ce{m(CO) = n \times M = 420.23 \times 28.01 = 11\ 771\ g}\)
\[\ce{[CO] = \frac{11\ 771}{100\ 000} = 0.118 g L^{-1}}\]
\(\ce{\text{Since} [CO] > 0.100 g L^{-1}, \text{it is dangerous to human health.}}\)
\(\ce{\text{Volume of garage} = 5 \times 5 \times 4 = 100 \text{m}^2}\)
\(\ce{100 \text{m}^2} = 100\ 000\ \text{L}\)
\[\ce{\text{n(octane)} = \frac{n}{M} = \frac{6000}{114.224} = 52.53 \text{moles}} \]
\(\ce{Molar ratio of Octane : CO = 1:8}\)
\(\ce{n(CO) = 8 \times 52.53 = 420.23}\)
\(\ce{m(CO) = n \times M = 420.23 \times 28.01 = 11\ 771\ g}\)
\[\ce{[CO] = \frac{11\ 771}{100\ 000} = 0.118 g L^{-1}}\]
\(\ce{\text{Since} [CO] > 0.100 g L^{-1}, \text{it is dangerous to human health.}}\)