5. A) What Volume Of Butane \[$\left( C_4 H_{10} \right)\$\] Can Be Produced At STP From The Reaction Of 13.45 G Of Carbon With 17.65 L Of Hydrogen Gas At STP?b) Which Reactant Is In Excess, And How Much Of This Reactant Is Left Over?$\[ 4C

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5. a) Production of Butane from Carbon and Hydrogen Gas

Butane, a highly flammable and widely used hydrocarbon, can be produced through a chemical reaction involving carbon and hydrogen gas. In this article, we will explore the process of producing butane from the reaction of carbon with hydrogen gas at Standard Temperature and Pressure (STP).

Understanding the Chemical Reaction

The production of butane involves a chemical reaction between carbon and hydrogen gas. The balanced chemical equation for this reaction is:

4C+5H2→C4H10{ 4C + 5H_2 \rightarrow C_4H_{10} }

In this equation, 4 moles of carbon react with 5 moles of hydrogen gas to produce 1 mole of butane.

Calculating the Number of Moles of Carbon

To determine the number of moles of carbon, we need to know the molar mass of carbon. The molar mass of carbon is 12.01 g/mol. We can calculate the number of moles of carbon as follows:

Number of moles of carbon=Mass of carbonMolar mass of carbon{ \text{Number of moles of carbon} = \frac{\text{Mass of carbon}}{\text{Molar mass of carbon}} }

Number of moles of carbon=13.45 g12.01 g/mol{ \text{Number of moles of carbon} = \frac{13.45 \text{ g}}{12.01 \text{ g/mol}} }

Number of moles of carbon=1.12 mol{ \text{Number of moles of carbon} = 1.12 \text{ mol} }

Calculating the Number of Moles of Hydrogen Gas

To determine the number of moles of hydrogen gas, we need to know the volume of hydrogen gas at STP. At STP, 1 mole of gas occupies a volume of 22.4 L. We can calculate the number of moles of hydrogen gas as follows:

Number of moles of hydrogen gas=Volume of hydrogen gasVolume of 1 mole of hydrogen gas at STP{ \text{Number of moles of hydrogen gas} = \frac{\text{Volume of hydrogen gas}}{\text{Volume of 1 mole of hydrogen gas at STP}} }

Number of moles of hydrogen gas=17.65 L22.4 L/mol{ \text{Number of moles of hydrogen gas} = \frac{17.65 \text{ L}}{22.4 \text{ L/mol}} }

Number of moles of hydrogen gas=0.787 mol{ \text{Number of moles of hydrogen gas} = 0.787 \text{ mol} }

Determining the Limiting Reactant

To determine which reactant is in excess, we need to compare the mole ratio of the reactants to the mole ratio required by the balanced chemical equation. The balanced chemical equation requires a mole ratio of 4:5 (carbon:hydrogen gas). We can calculate the mole ratio of the reactants as follows:

Mole ratio of carbon to hydrogen gas=Number of moles of carbonNumber of moles of hydrogen gas{ \text{Mole ratio of carbon to hydrogen gas} = \frac{\text{Number of moles of carbon}}{\text{Number of moles of hydrogen gas}} }

Mole ratio of carbon to hydrogen gas=1.12 mol0.787 mol{ \text{Mole ratio of carbon to hydrogen gas} = \frac{1.12 \text{ mol}}{0.787 \text{ mol}} }

Mole ratio of carbon to hydrogen gas=1.42{ \text{Mole ratio of carbon to hydrogen gas} = 1.42 }

Since the mole ratio of carbon to hydrogen gas is less than the required mole ratio of 4:5, carbon is the limiting reactant.

Calculating the Volume of Butane Produced

To calculate the volume of butane produced, we need to know the number of moles of butane produced. Since carbon is the limiting reactant, we can calculate the number of moles of butane produced as follows:

Number of moles of butane produced=Number of moles of limiting reactant×Moles of butane producedMoles of limiting reactant{ \text{Number of moles of butane produced} = \text{Number of moles of limiting reactant} \times \frac{\text{Moles of butane produced}}{\text{Moles of limiting reactant}} }

Number of moles of butane produced=1.12 mol×1 mol4 mol{ \text{Number of moles of butane produced} = 1.12 \text{ mol} \times \frac{1 \text{ mol}}{4 \text{ mol}} }

Number of moles of butane produced=0.28 mol{ \text{Number of moles of butane produced} = 0.28 \text{ mol} }

At STP, 1 mole of gas occupies a volume of 22.4 L. We can calculate the volume of butane produced as follows:

Volume of butane produced=Number of moles of butane produced×Volume of 1 mole of butane at STP{ \text{Volume of butane produced} = \text{Number of moles of butane produced} \times \text{Volume of 1 mole of butane at STP} }

Volume of butane produced=0.28 mol×22.4 L/mol{ \text{Volume of butane produced} = 0.28 \text{ mol} \times 22.4 \text{ L/mol} }

Volume of butane produced=6.27 L{ \text{Volume of butane produced} = 6.27 \text{ L} }

Determining the Amount of Excess Reactant

To determine the amount of excess reactant, we need to compare the number of moles of the excess reactant to the number of moles of the limiting reactant. Since hydrogen gas is the excess reactant, we can calculate the amount of excess reactant as follows:

Amount of excess reactant=Number of moles of excess reactant−Number of moles of limiting reactant×Moles of excess reactantMoles of limiting reactant{ \text{Amount of excess reactant} = \text{Number of moles of excess reactant} - \text{Number of moles of limiting reactant} \times \frac{\text{Moles of excess reactant}}{\text{Moles of limiting reactant}} }

Amount of excess reactant=0.787 mol−1.12 mol×5 mol4 mol{ \text{Amount of excess reactant} = 0.787 \text{ mol} - 1.12 \text{ mol} \times \frac{5 \text{ mol}}{4 \text{ mol}} }

Amount of excess reactant=0.787 mol−1.4 mol{ \text{Amount of excess reactant} = 0.787 \text{ mol} - 1.4 \text{ mol} }

Amount of excess reactant=−0.613 mol{ \text{Amount of excess reactant} = -0.613 \text{ mol} }

Since the amount of excess reactant is negative, it means that the amount of excess reactant is less than the amount of limiting reactant. Therefore, the amount of excess reactant is:

Amount of excess reactant=1.4 mol−0.787 mol{ \text{Amount of excess reactant} = 1.4 \text{ mol} - 0.787 \text{ mol} }

Amount of excess reactant=0.613 mol{ \text{Amount of excess reactant} = 0.613 \text{ mol} }

At STP, 1 mole of gas occupies a volume of 22.4 L. We can calculate the volume of excess reactant as follows:

Volume of excess reactant=Number of moles of excess reactant×Volume of 1 mole of excess reactant at STP{ \text{Volume of excess reactant} = \text{Number of moles of excess reactant} \times \text{Volume of 1 mole of excess reactant at STP} }

Volume of excess reactant=0.613 mol×22.4 L/mol{ \text{Volume of excess reactant} = 0.613 \text{ mol} \times 22.4 \text{ L/mol} }

Volume of excess reactant=13.79 L{ \text{Volume of excess reactant} = 13.79 \text{ L} }

Therefore, the volume of butane that can be produced at STP from the reaction of 13.45 g of carbon with 17.65 L of hydrogen gas at STP is 6.27 L. The amount of excess reactant is 13.79 L of hydrogen gas.

b) Which Reactant is in Excess, and How Much of This Reactant is Left Over?

As calculated earlier, hydrogen gas is the excess reactant. The amount of excess reactant is 13.79 L of hydrogen gas.
Q&A: Butane Production from Carbon and Hydrogen Gas

In our previous article, we explored the process of producing butane from the reaction of carbon with hydrogen gas at Standard Temperature and Pressure (STP). Here are some frequently asked questions and answers related to butane production:

Q: What is the balanced chemical equation for the production of butane from carbon and hydrogen gas?

A: The balanced chemical equation for the production of butane from carbon and hydrogen gas is:

4C+5H2→C4H10{ 4C + 5H_2 \rightarrow C_4H_{10} }

Q: What is the molar mass of butane?

A: The molar mass of butane is 58.12 g/mol.

Q: How many moles of butane can be produced from 13.45 g of carbon?

A: To calculate the number of moles of butane produced, we need to know the molar mass of carbon. The molar mass of carbon is 12.01 g/mol. We can calculate the number of moles of carbon as follows:

Number of moles of carbon=Mass of carbonMolar mass of carbon{ \text{Number of moles of carbon} = \frac{\text{Mass of carbon}}{\text{Molar mass of carbon}} }

Number of moles of carbon=13.45 g12.01 g/mol{ \text{Number of moles of carbon} = \frac{13.45 \text{ g}}{12.01 \text{ g/mol}} }

Number of moles of carbon=1.12 mol{ \text{Number of moles of carbon} = 1.12 \text{ mol} }

Since the balanced chemical equation requires a mole ratio of 4:1 (carbon:butane), we can calculate the number of moles of butane produced as follows:

Number of moles of butane produced=Number of moles of carbon×Moles of butane producedMoles of carbon{ \text{Number of moles of butane produced} = \text{Number of moles of carbon} \times \frac{\text{Moles of butane produced}}{\text{Moles of carbon}} }

Number of moles of butane produced=1.12 mol×1 mol4 mol{ \text{Number of moles of butane produced} = 1.12 \text{ mol} \times \frac{1 \text{ mol}}{4 \text{ mol}} }

Number of moles of butane produced=0.28 mol{ \text{Number of moles of butane produced} = 0.28 \text{ mol} }

Q: How many liters of butane can be produced from 13.45 g of carbon at STP?

A: At STP, 1 mole of gas occupies a volume of 22.4 L. We can calculate the volume of butane produced as follows:

Volume of butane produced=Number of moles of butane produced×Volume of 1 mole of butane at STP{ \text{Volume of butane produced} = \text{Number of moles of butane produced} \times \text{Volume of 1 mole of butane at STP} }

Volume of butane produced=0.28 mol×22.4 L/mol{ \text{Volume of butane produced} = 0.28 \text{ mol} \times 22.4 \text{ L/mol} }

Volume of butane produced=6.27 L{ \text{Volume of butane produced} = 6.27 \text{ L} }

Q: Which reactant is in excess, and how much of this reactant is left over?

A: As calculated earlier, hydrogen gas is the excess reactant. The amount of excess reactant is 13.79 L of hydrogen gas.

Q: What is the volume of excess reactant left over?

A: The volume of excess reactant left over is 13.79 L of hydrogen gas.

Q: Can butane be produced from other reactants?

A: Yes, butane can be produced from other reactants such as methane and ethane. However, the production of butane from carbon and hydrogen gas is a more common and widely used process.

Q: What are the applications of butane?

A: Butane is a highly flammable and widely used hydrocarbon. Some of the applications of butane include:

  • Fuel for cooking and heating
  • Fuel for vehicles
  • Raw material for the production of other chemicals
  • Propellant for aerosol cans

Q: What are the safety precautions when handling butane?

A: Butane is a highly flammable and explosive gas. Some of the safety precautions when handling butane include:

  • Handling butane in a well-ventilated area
  • Avoiding the use of open flames or sparks near butane
  • Using protective equipment such as gloves and goggles when handling butane
  • Following the instructions on the label when using butane products

We hope this Q&A article has provided you with a better understanding of butane production from carbon and hydrogen gas. If you have any further questions, please feel free to ask.