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A cylinder of gas contains \(11.6 \mathrm{~kg}\) of butane, if a normal family needs \(2.0 \times 10^{4} \mathrm{~kJ}\) of energy per day for cooking, how long will the cylinder last \(\left(\Delta H_{\text {combustian of bucane }}=-2600 \mathrm{~kJ} / \mathrm{mol}\right)\). (a) 52 days (b) 26 days (c) 116 days (d) none

Short Answer

Expert verified
(b) 26 days

Step by step solution

01

Determine the Molar Mass of Butane

Butane has a chemical formula of \( \text{C}_4\text{H}_{10} \). To find its molar mass, sum the atomic masses of its constituent atoms: Carbon has an atomic mass of approximately 12 g/mol, and Hydrogen has an atomic mass of approximately 1 g/mol. Therefore, the molar mass of Butane is: \[ \text{Molar Mass of C}_4\text{H}_{10} = 4 \times 12 + 10 \times 1 = 58 \text{ g/mol}. \] Thus, the molar mass is 58 g/mol or 0.058 kg/mol.
02

Calculate the Moles of Butane

To find the number of moles in 11.6 kg of butane, use the relationship between mass and moles: \[ \text{Moles of Butane} = \frac{\text{mass in kg}}{\text{molar mass in kg/mol}} = \frac{11.6 \text{ kg}}{0.058 \text{ kg/mol}}. \] After evaluating, this gives \[ \text{Moles of Butane} = 200 \text{ mol}. \]
03

Find the Total Energy from Combustion

The heat of combustion of butane is given as \(-2600 \text{ kJ/mol}\). Therefore, the total energy produced by 200 moles of butane is: \[ \text{Total Energy Produced} = 200 \text{ mol} \times (-2600 \text{ kJ/mol}) = -520,000 \text{ kJ}. \] The negative sign indicates that energy is released.
04

Calculate the Days the Cylinder Lasts

A family uses \(2.0 \times 10^4 \text{ kJ} \) per day. The number of days the cylinder will last is found by dividing the total energy available by the daily energy consumption: \[ \text{Days} = \frac{520,000 \text{ kJ}}{2.0 \times 10^4 \text{ kJ/day}} = 26 \text{ days}. \]
05

Conclusion and Answer

Based on the calculation, the cylinder will last for 26 days given the energy needs of the family. Therefore, the correct answer is \(\text{(b) 26 days}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass
Molar mass is a fundamental concept in chemistry that helps us understand the mass of a given element or compound per mole of its substance. For butane, which has the chemical formula \( \text{C}_4\text{H}_{10} \), the molar mass is calculated by summing the atomic masses of its constituent elements: carbon and hydrogen.

Butane consists of four carbon atoms and ten hydrogen atoms. We know from the periodic table that the atomic mass of carbon is approximately 12 g/mol, while that of hydrogen is about 1 g/mol. Using this information, we compute the molar mass of butane as follows:
  • Carbon: \( 4 \times 12 \text{ g/mol} = 48 \text{ g/mol} \)
  • Hydrogen: \( 10 \times 1 \text{ g/mol} = 10 \text{ g/mol} \)
Adding these together gives us a total molar mass of \( 58 \text{ g/mol} \). This is equivalent to \( 0.058 \text{ kg/mol} \) for calculations involving larger quantities, such as in gas cylinders.
Butane Combustion
Combustion is the process of burning a substance in the presence of oxygen to release energy. In the case of butane, the combustion reaction can be represented by the equation:

\[ 2 \text{C}_4\text{H}_{10} + 13 \text{O}_2 \rightarrow 8 \text{CO}_2 + 10 \text{H}_2\text{O} \]
This reaction releases a significant amount of energy in the form of heat. The enthalpy change for the combustion of butane is given as \(-2600 \text{ kJ/mol}\). The negative sign indicates that the process is exothermic, meaning energy is released as heat.

Understanding this energy release is crucial in energy calculations, especially when considering how much energy can be obtained from a given mass of butane. Efficient use of this energy is important for tasks such as cooking, heating, and even powering engines.
Energy Calculation
Energy calculation is essential when determining how much energy a given amount of butane can provide. Knowing the combustion energy per mole, we can calculate the total energy from a given mass of butane by first converting the mass to moles using the molar mass.

For example, with 200 moles of butane available, and each mole releasing \(-2600 \text{ kJ}\), the total energy released is:
  • \( 200 \text{ mol} \times -2600 \text{ kJ/mol} = -520,000 \text{ kJ} \)
The negative sign signifies that this energy is released. Such calculations are vital for practical applications, such as estimating how long a butane cylinder will last based on household energy consumption needs.
Gas Cylinders
Gas cylinders are a common storage solution for gases under high pressure, including butane. They are essential in household cooking and other energy applications due to their portability and efficiency in storing gases.

When we talk about a cylinder containing 11.6 kg of butane, we're referring to the total mass of butane stored. To estimate how long such a cylinder can last for a typical family, we must assess its energy output potential. This involves converting the mass to moles, calculating the energy each mole can produce, and then determining how this energy matches the family's daily energy requirements.

Efficient cylinder usage ensures families can optimize their budget and resource management, making these calculations practical and essential.
Stoichiometry
Stoichiometry is the branch of chemistry that involves the calculation of reactants and products in chemical reactions. It is crucial for understanding the ratios in which substances react and the amounts needed or produced.

In our butane combustion example, stoichiometry allows us to balance the equation and ensure all elements are accounted for on both reactant and product sides:

\[ 2 \text{C}_4\text{H}_{10} + 13 \text{O}_2 \rightarrow 8 \text{CO}_2 + 10 \text{H}_2\text{O} \]
This balanced equation tells us that 2 moles of butane require 13 moles of oxygen to produce 8 moles of carbon dioxide and 10 moles of water.

Stoichiometry is not only about balancing equations but also helps in calculating how long a gas cylinder will last, by understanding how much energy results from these reactions. It's a key part of optimizing the use of chemical substances in everyday contexts.

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Most popular questions from this chapter

A person requires \(2870 \mathrm{kcal}\) of energy to lead normal daily life. If heat of combustion of cane sugar is \(-\) \(1349 \mathrm{kcal}\), then his daily consumption of sugar is (a) \(728 \mathrm{~g}\) (b) \(0.728 \mathrm{~g}\) (c) \(342 \mathrm{~g}\) (d) \(0.342 \mathrm{~g}\)

Carnot's cycle operates between \(\mathrm{T}\) and \(300 \mathrm{~K}\) (T > 300K). It is \(25 \%\) efficient when temperature of die source (T) is (a) \(350 \mathrm{~K}\) (b) \(400 \mathrm{~K}\) (c) \(450 \mathrm{~K}\) (d) \(375 \mathrm{~K}\)

The combustion of \(2.650 \mathrm{~g}\) of indium raised the temperature of a calorimeter \(1.055 \mathrm{C}\). The energy equivalent of the calorimeter was \(2412.81 \mathrm{cal} /{ }^{\circ} \mathrm{C}\). Calculate the molar enthalpy combustion of indium (Molar mass of indium \(114.8\) ) (a) \(2412.81 \mathrm{cal} / \mathrm{mole}\) (b) \(119683.8 \mathrm{cal} / \mathrm{mole}\) (c) \(110,273.6 \mathrm{cal} / \mathrm{mole}\) (d) None

Internal energy and pressure of a gas of unit volume are related as: (a) \(\mathrm{P}=\frac{2}{3} \mathrm{E}\) (b) \(\mathrm{P}=\frac{3}{2} \mathrm{E}\) (c) \(\mathrm{P}=\frac{\mathrm{E}}{2}\) (d) \(P=2 E\)

Calculate the mass of mercury which can be liberated from \(\mathrm{HgO}\) at \(25^{\circ} \mathrm{C}\) by the treatment of excess \(\mathrm{HgO}\) with \(10 \mathrm{kcal}\) of heat. Standard enthalpy of formation of \(\mathrm{Hg} \mathrm{O}\) is \(21.7 \mathrm{kcal} /\) mole (a) \(92.4 \mathrm{~g}\) (b) \(9.24 \mathrm{~g}\) (c) \(924 \mathrm{~g}\) (d) \(200 \mathrm{~g}\)

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