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100cm³ of hydrogen has sparked with 40cm³ of oxygen at 100 degree Celsius and 1atm. The total volume of the residual gases is what?

Feb 20

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Feb 20

To solve this problem, we first need to determine the moles of each gas present before the reaction occurs.

Using the ideal gas law equation, we can calculate the number of moles of hydrogen and oxygen:

n = PV/RT

For hydrogen:

n(H2) = (1 atm)(100 cm³) / (0.0821 L.atm/mol.K)(373 K)

n(H2) = 0.270 moles

For oxygen:

n(O2) = (1 atm)(40 cm³) / (0.0821 L.atm/mol.K)(373 K)

n(O2) = 0.108 moles

Since the reaction between hydrogen and oxygen is a 2:1 ratio, all the oxygen will be consumed and 0.054 moles of hydrogen will be consumed. This will produce 0.054 moles of water vapor.

Now we need to calculate the volume of the residual gases after the reaction. The total volume of the gases before the reaction was 140 cm³. After the reaction, the volume of water vapor produced will be 0.054 moles * 22.4 L/mol = 1.21 L.

Therefore, the total volume of the residual gases will be:

140 cm³ - 1.21 L = 138.79 cm³

So, the total volume of the residual gases is 138.79 cm³.

Using the ideal gas law equation, we can calculate the number of moles of hydrogen and oxygen:

n = PV/RT

For hydrogen:

n(H2) = (1 atm)(100 cm³) / (0.0821 L.atm/mol.K)(373 K)

n(H2) = 0.270 moles

For oxygen:

n(O2) = (1 atm)(40 cm³) / (0.0821 L.atm/mol.K)(373 K)

n(O2) = 0.108 moles

Since the reaction between hydrogen and oxygen is a 2:1 ratio, all the oxygen will be consumed and 0.054 moles of hydrogen will be consumed. This will produce 0.054 moles of water vapor.

Now we need to calculate the volume of the residual gases after the reaction. The total volume of the gases before the reaction was 140 cm³. After the reaction, the volume of water vapor produced will be 0.054 moles * 22.4 L/mol = 1.21 L.

Therefore, the total volume of the residual gases will be:

140 cm³ - 1.21 L = 138.79 cm³

So, the total volume of the residual gases is 138.79 cm³.