An "ideal" fuel'air mixture in which both the fuel and the oxygen in the air are completely consumed is called the "stoichiometric" mixture. The stoichiometric mixture is usually specified as the ratio of the mass of air to the mass of a particular fuel. For example, if a fuel (for example, natural gas, CH4) required 17.2 kg of air to precisely burn 1 kg of the fuel (that is, all the fuel burned and there was no oxygen left in the air afterwards) then the stoichiometric ratio for that fuel would be 17.2: 1. Calculate the partial pressures of the natural gas and oxygen in the mixture based on the stoichiometric ratio given above at 293.15 K. The volume of the internal combustion engine is considered to be 2.500 L. The mass of the fuel is 160.0 g. Mass fraction of oxygen in air is 23.1%. Molar mass of natural gas is 16.043 g/mol. P natural gas = atm Poxygen=[ atm

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An "ideal" fuel/air mixture in which both the fuel and the oxygen in the air are completely consumed is called the "stoichiometric" mixture. The stoichiometric mixture is usually specified as the ratio of the mass of air to the mass of a particular fuel. For example, if a fuel (for
example, natural gas, CH4) required 17.2 kg of air to precisely burn 1 kg of the fuel (that is, all the fuel burned and there was no oxygen left in the air afterwards) then the stoichiometric ratio for that fuel would be 17.2: 1.
Calculate the partial pressures of the natural gas and oxygen in the mixture based on the stoichiometric ratio given above at 293.15 K. The volume of the internal combustion engine is considered to be 2.500 L. The mass of the fuel is 160.0 g.
Mass fraction of oxygen in air is 23.1%.
Molar mass of natural gas is 16.043 g/mol.
atm
natural gas =
Poxygen
atm
Transcribed Image Text:An "ideal" fuel/air mixture in which both the fuel and the oxygen in the air are completely consumed is called the "stoichiometric" mixture. The stoichiometric mixture is usually specified as the ratio of the mass of air to the mass of a particular fuel. For example, if a fuel (for example, natural gas, CH4) required 17.2 kg of air to precisely burn 1 kg of the fuel (that is, all the fuel burned and there was no oxygen left in the air afterwards) then the stoichiometric ratio for that fuel would be 17.2: 1. Calculate the partial pressures of the natural gas and oxygen in the mixture based on the stoichiometric ratio given above at 293.15 K. The volume of the internal combustion engine is considered to be 2.500 L. The mass of the fuel is 160.0 g. Mass fraction of oxygen in air is 23.1%. Molar mass of natural gas is 16.043 g/mol. atm natural gas = Poxygen atm
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