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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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 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)
\(\ce{\text{Volume of garage} = 5 \times 5 \times 4 = 100 \text{m}^2}\)
\(\ce{100 \text{m}^2} = 100\ 000\ \text{L}\)
\[\ce{\text{n(octane)} = \frac{n}{M} = \frac{6000}{114.224} = 52.53 \text{moles}} \]
\(\ce{Molar ratio of Octane : CO = 1:8}\)
\(\ce{n(CO) = 8 \times 52.53 = 420.23}\)
\(\ce{m(CO) = n \times M = 420.23 \times 28.01 = 11\ 771\ g}\)
\[\ce{[CO] = \frac{11\ 771}{100\ 000} = 0.118 g L^{-1}}\]
\(\ce{\text{Since} [CO] > 0.100 g L^{-1}, \text{it is dangerous to human health.}}\)
\(\ce{\text{Volume of garage} = 5 \times 5 \times 4 = 100 \text{m}^2}\)
\(\ce{100 \text{m}^2} = 100\ 000\ \text{L}\)
\[\ce{\text{n(octane)} = \frac{n}{M} = \frac{6000}{114.224} = 52.53 \text{moles}} \]
\(\ce{Molar ratio of Octane : CO = 1:8}\)
\(\ce{n(CO) = 8 \times 52.53 = 420.23}\)
\(\ce{m(CO) = n \times M = 420.23 \times 28.01 = 11\ 771\ g}\)
\[\ce{[CO] = \frac{11\ 771}{100\ 000} = 0.118 g L^{-1}}\]
\(\ce{\text{Since} [CO] > 0.100 g L^{-1}, \text{it is dangerous to human health.}}\)
What volume of carbon dioxide will be produced if 10.3 g of glucose is fermented at 25°C and 100 kPa?
`D`
\(\ce{C6H12O6 -> 2CO2 + 2C2H5OH}\)
\[\ce{n(C6H12O6) = \frac{10.3}{6 \times 12.01 + 12 \times 1.008 + 6 \times 16} = \frac{10.3}{180.156} = 0.057 mol}\]
\(\ce{n(CO2) = 2 \times n(C6H12O6) = 0.114 mol}\)
\(\ce{\text{Volume} (CO2) = 0.114 \times 24.79 = 2.83 L}\)
`=>D`