when 3g of C2H6 is completely burnt then find out produced volume of CO2 at STP.
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Answer:
As per given chemical equation 2 moles of Ethane requires 7 moles of oxygen for complete combustion
As per given chemical equation 2 moles of Ethane requires 7 moles of oxygen for complete combustionNow 1 mole of a substance occupies 22.4 L of volume at STP so 2 moles of ethane occupies 44.8 L volume and 7 moles of oxygen occupies 156.8 L volume
As per given chemical equation 2 moles of Ethane requires 7 moles of oxygen for complete combustionNow 1 mole of a substance occupies 22.4 L of volume at STP so 2 moles of ethane occupies 44.8 L volume and 7 moles of oxygen occupies 156.8 L volumeBut its given that available oxygen is 21 L i.e moles of oxygen available are 21/(7×22.4) = 3/22.4
As per given chemical equation 2 moles of Ethane requires 7 moles of oxygen for complete combustionNow 1 mole of a substance occupies 22.4 L of volume at STP so 2 moles of ethane occupies 44.8 L volume and 7 moles of oxygen occupies 156.8 L volumeBut its given that available oxygen is 21 L i.e moles of oxygen available are 21/(7×22.4) = 3/22.4Now moles of Ethane that will react with 3/22.4 moles of oxygen are (2×3)/(7×22.4) moles i.e 6/7×22.4 moles
As per given chemical equation 2 moles of Ethane requires 7 moles of oxygen for complete combustionNow 1 mole of a substance occupies 22.4 L of volume at STP so 2 moles of ethane occupies 44.8 L volume and 7 moles of oxygen occupies 156.8 L volumeBut its given that available oxygen is 21 L i.e moles of oxygen available are 21/(7×22.4) = 3/22.4Now moles of Ethane that will react with 3/22.4 moles of oxygen are (2×3)/(7×22.4) moles i.e 6/7×22.4 molesNow volume of 1 mole of substance as stated above is 22.4 L hence volume of substance required to react with 21 L of oxygen is 6/(7×22.4) ×22.4 ==6/7 = 0.8571 litres of Ethane to be approximate
As per given chemical equation 2 moles of Ethane requires 7 moles of oxygen for complete combustionNow 1 mole of a substance occupies 22.4 L of volume at STP so 2 moles of ethane occupies 44.8 L volume and 7 moles of oxygen occupies 156.8 L volumeBut its given that available oxygen is 21 L i.e moles of oxygen available are 21/(7×22.4) = 3/22.4Now moles of Ethane that will react with 3/22.4 moles of oxygen are (2×3)/(7×22.4) moles i.e 6/7×22.4 molesNow volume of 1 mole of substance as stated above is 22.4 L hence volume of substance required to react with 21 L of oxygen is 6/(7×22.4) ×22.4 ==6/7 = 0.8571 litres of Ethane to be approximateThat is a detailed answer
As per given chemical equation 2 moles of Ethane requires 7 moles of oxygen for complete combustionNow 1 mole of a substance occupies 22.4 L of volume at STP so 2 moles of ethane occupies 44.8 L volume and 7 moles of oxygen occupies 156.8 L volumeBut its given that available oxygen is 21 L i.e moles of oxygen available are 21/(7×22.4) = 3/22.4Now moles of Ethane that will react with 3/22.4 moles of oxygen are (2×3)/(7×22.4) moles i.e 6/7×22.4 molesNow volume of 1 mole of substance as stated above is 22.4 L hence volume of substance required to react with 21 L of oxygen is 6/(7×22.4) ×22.4 ==6/7 = 0.8571 litres of Ethane to be approximateThat is a detailed answerHope this helps !!!
As per given chemical equation 2 moles of Ethane requires 7 moles of oxygen for complete combustionNow 1 mole of a substance occupies 22.4 L of volume at STP so 2 moles of ethane occupies 44.8 L volume and 7 moles of oxygen occupies 156.8 L volumeBut its given that available oxygen is 21 L i.e moles of oxygen available are 21/(7×22.4) = 3/22.4Now moles of Ethane that will react with 3/22.4 moles of oxygen are (2×3)/(7×22.4) moles i.e 6/7×22.4 molesNow volume of 1 mole of substance as stated above is 22.4 L hence volume of substance required to react with 21 L of oxygen is 6/(7×22.4) ×22.4 ==6/7 = 0.8571 litres of Ethane to be approximateThat is a detailed answerHope this helps !!!Thank you .