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At STP volume of an ideal gas is 22.4L/mol.
Volume of ethane mixed = `16L`
Amount of Ethane mixed = `16/22.4`
`C_2H_6:CO_2 = 2:4 = 1:2`
Amount of `CO_2` produced = `(16/22.4)*2`
Volume of `CO_2` produced = `(16/22.4)*2*22.4 = 32 L`
So there will be 32L of `CO_2` will be produced.
`CO_2` and `C_2H_6` act as ideal gasses at the given conditions.
Balanced chemical equation:
2C2H6 + 7O2 --> 4 CO2 + 6H2O
2 moles of ethane produce 4 moles of CO2.
Given volume of ethane gas = 16 lt.
At STP, 1 mole of a gas = 22.4 lt of gas
In the present case, 2 moles of ethane = 2x22.4 = 44.8 lts of ethane produces 4 x 22.4 lts (= 89.6 lts) of CO2
therefore, using unitary method,
16 l of ethane will produce = 89.6 x 16/44.8 lts of CO2 = 32 lts of CO2.
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