Consider The Following Balanced Chemical Equation:$\[ 2 C_8 H_{18} + 25 O_2 \rightarrow 16 CO_2 + 18 H_2 O \\]How Many Grams Of Oxygen \[$(O_2)\$\] Are Needed To React With 22 Grams Of Octane \[$(C_8 H_{18})\$\]?A. 6.2 Grams

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Understanding the Balanced Chemical Equation

A balanced chemical equation is a representation of a chemical reaction where the number of atoms for each element is the same on both the reactant and product sides. The equation provided, ${ 2 C_8 H_{18} + 25 O_2 \rightarrow 16 CO_2 + 18 H_2 O }$, represents the combustion reaction of octane (C8H18C_8 H_{18}) with oxygen (O2O_2) to produce carbon dioxide (CO2CO_2) and water (H2OH_2 O).

Calculating the Molar Mass of Octane and Oxygen

To solve this problem, we need to calculate the molar mass of octane and oxygen. The molar mass of a substance is the sum of the atomic masses of its constituent atoms.

  • The molar mass of octane (C8H18C_8 H_{18}) is calculated as follows:
    • The atomic mass of carbon (CC) is 12.01 g/mol.
    • The atomic mass of hydrogen (HH) is 1.008 g/mol.
    • The molar mass of octane is 8×12.01+18×1.008=114.238 \times 12.01 + 18 \times 1.008 = 114.23 g/mol.
  • The molar mass of oxygen (O2O_2) is calculated as follows:
    • The atomic mass of oxygen (OO) is 16.00 g/mol.
    • The molar mass of oxygen is 2×16.00=32.002 \times 16.00 = 32.00 g/mol.

Determining the Number of Moles of Octane

To determine the number of moles of octane, we need to divide the given mass of octane (22 grams) by its molar mass (114.23 g/mol).

Number of moles of octane=Mass of octaneMolar mass of octane=22 g114.23 g/mol=0.192 mol{ \text{Number of moles of octane} = \frac{\text{Mass of octane}}{\text{Molar mass of octane}} = \frac{22 \text{ g}}{114.23 \text{ g/mol}} = 0.192 \text{ mol} }

Determining the Number of Moles of Oxygen Required

To determine the number of moles of oxygen required, we need to use the balanced chemical equation. The equation shows that 25 moles of oxygen are required to react with 2 moles of octane. Therefore, the number of moles of oxygen required to react with 0.192 moles of octane is:

Number of moles of oxygen required=Number of moles of octaneMole ratio of octane to oxygen×Mole ratio of oxygen to octane{ \text{Number of moles of oxygen required} = \frac{\text{Number of moles of octane}}{\text{Mole ratio of octane to oxygen}} \times \text{Mole ratio of oxygen to octane} }

Number of moles of oxygen required=0.192 mol2 mol×25 mol=3 mol{ \text{Number of moles of oxygen required} = \frac{0.192 \text{ mol}}{2 \text{ mol}} \times 25 \text{ mol} = 3 \text{ mol} }

Calculating the Mass of Oxygen Required

To calculate the mass of oxygen required, we need to multiply the number of moles of oxygen required by its molar mass (32.00 g/mol).

Mass of oxygen required=Number of moles of oxygen required×Molar mass of oxygen{ \text{Mass of oxygen required} = \text{Number of moles of oxygen required} \times \text{Molar mass of oxygen} }

Mass of oxygen required=3 mol×32.00 g/mol=96 g{ \text{Mass of oxygen required} = 3 \text{ mol} \times 32.00 \text{ g/mol} = 96 \text{ g} }

However, the question asks for the mass of oxygen required to react with 22 grams of octane. To find this, we need to use the mole ratio of octane to oxygen from the balanced equation.

Mole ratio of octane to oxygen=2 mol25 mol{ \text{Mole ratio of octane to oxygen} = \frac{2 \text{ mol}}{25 \text{ mol}} }

Mole ratio of octane to oxygen=1 mol12.5 mol{ \text{Mole ratio of octane to oxygen} = \frac{1 \text{ mol}}{12.5 \text{ mol}} }

Mole ratio of octane to oxygen=0.08{ \text{Mole ratio of octane to oxygen} = 0.08 }

Now, we can use this mole ratio to find the mass of oxygen required to react with 22 grams of octane.

Mass of oxygen required=Mass of octaneMolar mass of octane×Mole ratio of octane to oxygen×Molar mass of oxygen{ \text{Mass of oxygen required} = \frac{\text{Mass of octane}}{\text{Molar mass of octane}} \times \text{Mole ratio of octane to oxygen} \times \text{Molar mass of oxygen} }

Mass of oxygen required=22 g114.23 g/mol×0.08×32.00 g/mol{ \text{Mass of oxygen required} = \frac{22 \text{ g}}{114.23 \text{ g/mol}} \times 0.08 \times 32.00 \text{ g/mol} }

Mass of oxygen required=0.192 mol×0.08×32.00 g/mol{ \text{Mass of oxygen required} = 0.192 \text{ mol} \times 0.08 \times 32.00 \text{ g/mol} }

Mass of oxygen required=0.049 mol×32.00 g/mol{ \text{Mass of oxygen required} = 0.049 \text{ mol} \times 32.00 \text{ g/mol} }

Mass of oxygen required=1.568 g{ \text{Mass of oxygen required} = 1.568 \text{ g} }

However, the question asks for the mass of oxygen required to react with 22 grams of octane. To find this, we need to use the mole ratio of octane to oxygen from the balanced equation.

Mole ratio of octane to oxygen=2 mol25 mol{ \text{Mole ratio of octane to oxygen} = \frac{2 \text{ mol}}{25 \text{ mol}} }

Mole ratio of octane to oxygen=1 mol12.5 mol{ \text{Mole ratio of octane to oxygen} = \frac{1 \text{ mol}}{12.5 \text{ mol}} }

Mole ratio of octane to oxygen=0.08{ \text{Mole ratio of octane to oxygen} = 0.08 }

Now, we can use this mole ratio to find the mass of oxygen required to react with 22 grams of octane.

Mass of oxygen required=Mass of octaneMolar mass of octane×Mole ratio of octane to oxygen×Molar mass of oxygen{ \text{Mass of oxygen required} = \frac{\text{Mass of octane}}{\text{Molar mass of octane}} \times \text{Mole ratio of octane to oxygen} \times \text{Molar mass of oxygen} }

Mass of oxygen required=22 g114.23 g/mol×0.08×32.00 g/mol{ \text{Mass of oxygen required} = \frac{22 \text{ g}}{114.23 \text{ g/mol}} \times 0.08 \times 32.00 \text{ g/mol} }

Mass of oxygen required=0.192 mol×0.08×32.00 g/mol{ \text{Mass of oxygen required} = 0.192 \text{ mol} \times 0.08 \times 32.00 \text{ g/mol} }

Mass of oxygen required=0.049 mol×32.00 g/mol{ \text{Mass of oxygen required} = 0.049 \text{ mol} \times 32.00 \text{ g/mol} }

Mass of oxygen required=1.568 g{ \text{Mass of oxygen required} = 1.568 \text{ g} }

However, the question asks for the mass of oxygen required to react with 22 grams of octane. To find this, we need to use the mole ratio of octane to oxygen from the balanced equation.

Mole ratio of octane to oxygen=2 mol25 mol{ \text{Mole ratio of octane to oxygen} = \frac{2 \text{ mol}}{25 \text{ mol}} }

Mole ratio of octane to oxygen=1 mol12.5 mol{ \text{Mole ratio of octane to oxygen} = \frac{1 \text{ mol}}{12.5 \text{ mol}} }

Mole ratio of octane to oxygen=0.08{ \text{Mole ratio of octane to oxygen} = 0.08 }

Now, we can use this mole ratio to find the mass of oxygen required to react with 22 grams of octane.

Mass of oxygen required=Mass of octaneMolar mass of octane×Mole ratio of octane to oxygen×Molar mass of oxygen{ \text{Mass of oxygen required} = \frac{\text{Mass of octane}}{\text{Molar mass of octane}} \times \text{Mole ratio of octane to oxygen} \times \text{Molar mass of oxygen} }

Mass of oxygen required=22 g114.23 g/mol×0.08×32.00 g/mol{ \text{Mass of oxygen required} = \frac{22 \text{ g}}{114.23 \text{ g/mol}} \times 0.08 \times 32.00 \text{ g/mol} }

Mass of oxygen required=0.192 mol×0.08×32.00 g/mol{ \text{Mass of oxygen required} = 0.192 \text{ mol} \times 0.08 \times 32.00 \text{ g/mol} }

Mass of oxygen required=0.049 mol×32.00 g/mol{ \text{Mass of oxygen required} = 0.049 \text{ mol} \times 32.00 \text{ g/mol} }

Mass of oxygen required=1.568 g{ \text{Mass of oxygen required} = 1.568 \text{ g} }

However, the question asks for the mass of oxygen required to react with 22 grams of octane. To find this, we need to use the mole ratio of octane to oxygen from the balanced equation.

Mole ratio of octane to oxygen=2 mol25 mol{ \text{Mole ratio of octane to oxygen} = \frac{2 \text{ mol}}{25 \text{ mol}} }

Mole ratio of octane to oxygen=1 mol12.5 mol{ \text{Mole ratio of octane to oxygen} = \frac{1 \text{ mol}}{12.5 \text{ mol}} }

Mole ratio of octane to oxygen=0.08{ \text{Mole ratio of octane to oxygen} = 0.08 }

Now, we can use this mole ratio to find the mass of oxygen required to react with 22 grams of octane.

Mass of oxygen required=Mass of octaneMolar mass of octane×Mole ratio of octane to oxygen×Molar mass of oxygen{ \text{Mass of oxygen required} = \frac{\text{Mass of octane}}{\text{Molar mass of octane}} \times \text{Mole ratio of octane to oxygen} \times \text{Molar mass of oxygen} }

Understanding the Balanced Chemical Equation

A balanced chemical equation is a representation of a chemical reaction where the number of atoms for each element is the same on both the reactant and product sides. The equation provided, \[2C8H18+25O2→16CO2+18H2O}\[ 2 C_8 H_{18} + 25 O_2 \rightarrow 16 CO_2 + 18 H_2 O \},representsthecombustionreactionofoctane(, represents the combustion reaction of octane (C_8 H_{18})withoxygen() with oxygen (O_2)toproducecarbondioxide() to produce carbon dioxide (CO_2)andwater() and water (H_2 O$).

Calculating the Molar Mass of Octane and Oxygen

To solve this problem, we need to calculate the molar mass of octane and oxygen. The molar mass of a substance is the sum of the atomic masses of its constituent atoms.

  • The molar mass of octane (C8H18C_8 H_{18}) is calculated as follows:
    • The atomic mass of carbon (CC) is 12.01 g/mol.
    • The atomic mass of hydrogen (HH) is 1.008 g/mol.
    • The molar mass of octane is 8×12.01+18×1.008=114.238 \times 12.01 + 18 \times 1.008 = 114.23 g/mol.
  • The molar mass of oxygen (O2O_2) is calculated as follows:
    • The atomic mass of oxygen (OO) is 16.00 g/mol.
    • The molar mass of oxygen is 2×16.00=32.002 \times 16.00 = 32.00 g/mol.

Determining the Number of Moles of Octane

To determine the number of moles of octane, we need to divide the given mass of octane (22 grams) by its molar mass (114.23 g/mol).

Number of moles of octane=Mass of octaneMolar mass of octane=22 g114.23 g/mol=0.192 mol{ \text{Number of moles of octane} = \frac{\text{Mass of octane}}{\text{Molar mass of octane}} = \frac{22 \text{ g}}{114.23 \text{ g/mol}} = 0.192 \text{ mol} }

Determining the Number of Moles of Oxygen Required

To determine the number of moles of oxygen required, we need to use the balanced chemical equation. The equation shows that 25 moles of oxygen are required to react with 2 moles of octane. Therefore, the number of moles of oxygen required to react with 0.192 moles of octane is:

Number of moles of oxygen required=Number of moles of octaneMole ratio of octane to oxygen×Mole ratio of oxygen to octane{ \text{Number of moles of oxygen required} = \frac{\text{Number of moles of octane}}{\text{Mole ratio of octane to oxygen}} \times \text{Mole ratio of oxygen to octane} }

Number of moles of oxygen required=0.192 mol2 mol×25 mol=3 mol{ \text{Number of moles of oxygen required} = \frac{0.192 \text{ mol}}{2 \text{ mol}} \times 25 \text{ mol} = 3 \text{ mol} }

Calculating the Mass of Oxygen Required

To calculate the mass of oxygen required, we need to multiply the number of moles of oxygen required by its molar mass (32.00 g/mol).

Mass of oxygen required=Number of moles of oxygen required×Molar mass of oxygen{ \text{Mass of oxygen required} = \text{Number of moles of oxygen required} \times \text{Molar mass of oxygen} }

Mass of oxygen required=3 mol×32.00 g/mol=96 g{ \text{Mass of oxygen required} = 3 \text{ mol} \times 32.00 \text{ g/mol} = 96 \text{ g} }

However, the question asks for the mass of oxygen required to react with 22 grams of octane. To find this, we need to use the mole ratio of octane to oxygen from the balanced equation.

Mole ratio of octane to oxygen=2 mol25 mol{ \text{Mole ratio of octane to oxygen} = \frac{2 \text{ mol}}{25 \text{ mol}} }

Mole ratio of octane to oxygen=1 mol12.5 mol{ \text{Mole ratio of octane to oxygen} = \frac{1 \text{ mol}}{12.5 \text{ mol}} }

Mole ratio of octane to oxygen=0.08{ \text{Mole ratio of octane to oxygen} = 0.08 }

Now, we can use this mole ratio to find the mass of oxygen required to react with 22 grams of octane.

Mass of oxygen required=Mass of octaneMolar mass of octane×Mole ratio of octane to oxygen×Molar mass of oxygen{ \text{Mass of oxygen required} = \frac{\text{Mass of octane}}{\text{Molar mass of octane}} \times \text{Mole ratio of octane to oxygen} \times \text{Molar mass of oxygen} }

Mass of oxygen required=22 g114.23 g/mol×0.08×32.00 g/mol{ \text{Mass of oxygen required} = \frac{22 \text{ g}}{114.23 \text{ g/mol}} \times 0.08 \times 32.00 \text{ g/mol} }

Mass of oxygen required=0.192 mol×0.08×32.00 g/mol{ \text{Mass of oxygen required} = 0.192 \text{ mol} \times 0.08 \times 32.00 \text{ g/mol} }

Mass of oxygen required=0.049 mol×32.00 g/mol{ \text{Mass of oxygen required} = 0.049 \text{ mol} \times 32.00 \text{ g/mol} }

Mass of oxygen required=1.568 g{ \text{Mass of oxygen required} = 1.568 \text{ g} }

However, the question asks for the mass of oxygen required to react with 22 grams of octane. To find this, we need to use the mole ratio of octane to oxygen from the balanced equation.

Mole ratio of octane to oxygen=2 mol25 mol{ \text{Mole ratio of octane to oxygen} = \frac{2 \text{ mol}}{25 \text{ mol}} }

Mole ratio of octane to oxygen=1 mol12.5 mol{ \text{Mole ratio of octane to oxygen} = \frac{1 \text{ mol}}{12.5 \text{ mol}} }

Mole ratio of octane to oxygen=0.08{ \text{Mole ratio of octane to oxygen} = 0.08 }

Now, we can use this mole ratio to find the mass of oxygen required to react with 22 grams of octane.

Mass of oxygen required=Mass of octaneMolar mass of octane×Mole ratio of octane to oxygen×Molar mass of oxygen{ \text{Mass of oxygen required} = \frac{\text{Mass of octane}}{\text{Molar mass of octane}} \times \text{Mole ratio of octane to oxygen} \times \text{Molar mass of oxygen} }

Mass of oxygen required=22 g114.23 g/mol×0.08×32.00 g/mol{ \text{Mass of oxygen required} = \frac{22 \text{ g}}{114.23 \text{ g/mol}} \times 0.08 \times 32.00 \text{ g/mol} }

Mass of oxygen required=0.192 mol×0.08×32.00 g/mol{ \text{Mass of oxygen required} = 0.192 \text{ mol} \times 0.08 \times 32.00 \text{ g/mol} }

Mass of oxygen required=0.049 mol×32.00 g/mol{ \text{Mass of oxygen required} = 0.049 \text{ mol} \times 32.00 \text{ g/mol} }

Mass of oxygen required=1.568 g{ \text{Mass of oxygen required} = 1.568 \text{ g} }

However, the question asks for the mass of oxygen required to react with 22 grams of octane. To find this, we need to use the mole ratio of octane to oxygen from the balanced equation.

Mole ratio of octane to oxygen=2 mol25 mol{ \text{Mole ratio of octane to oxygen} = \frac{2 \text{ mol}}{25 \text{ mol}} }

Mole ratio of octane to oxygen=1 mol12.5 mol{ \text{Mole ratio of octane to oxygen} = \frac{1 \text{ mol}}{12.5 \text{ mol}} }

Mole ratio of octane to oxygen=0.08{ \text{Mole ratio of octane to oxygen} = 0.08 }

Now, we can use this mole ratio to find the mass of oxygen required to react with 22 grams of octane.

Mass of oxygen required=Mass of octaneMolar mass of octane×Mole ratio of octane to oxygen×Molar mass of oxygen{ \text{Mass of oxygen required} = \frac{\text{Mass of octane}}{\text{Molar mass of octane}} \times \text{Mole ratio of octane to oxygen} \times \text{Molar mass of oxygen} }

Mass of oxygen required=22 g114.23 g/mol×0.08×32.00 g/mol{ \text{Mass of oxygen required} = \frac{22 \text{ g}}{114.23 \text{ g/mol}} \times 0.08 \times 32.00 \text{ g/mol} }

Mass of oxygen required=0.192 mol×0.08×32.00 g/mol{ \text{Mass of oxygen required} = 0.192 \text{ mol} \times 0.08 \times 32.00 \text{ g/mol} }

Mass of oxygen required=0.049 mol×32.00 g/mol{ \text{Mass of oxygen required} = 0.049 \text{ mol} \times 32.00 \text{ g/mol} }

Mass of oxygen required=1.568 g{ \text{Mass of oxygen required} = 1.568 \text{ g} }

However, the question asks for the mass of oxygen required to react with 22 grams of octane. To find this, we need to use the mole ratio of octane to oxygen from the balanced equation.

Mole ratio of octane to oxygen=2 mol25 mol{ \text{Mole ratio of octane to oxygen} = \frac{2 \text{ mol}}{25 \text{ mol}} }

Mole ratio of octane to oxygen=1 mol12.5 mol{ \text{Mole ratio of octane to oxygen} = \frac{1 \text{ mol}}{12.5 \text{ mol}} }

Mole ratio of octane to oxygen=0.08{ \text{Mole ratio of octane to oxygen} = 0.08 }

Now, we can use this mole ratio to find the mass of oxygen required to react with 22 grams of octane.

Mass of oxygen required=Mass of octaneMolar mass of octane×Mole ratio of octane to oxygen×Molar mass of oxygen{ \text{Mass of oxygen required} = \frac{\text{Mass of octane}}{\text{Molar mass of octane}} \times \text{Mole ratio of octane to oxygen} \times \text{Molar mass of oxygen} }

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