6g C is burnt completely in excess of air. The volume CO₂ formed at STP is
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C+02-->CO2
12g Co2 reacts with 32 g 02 to produce 44gCO2
6g will react with 16g 02 to give 22g CO2
44gCO2 contains 22.4L 22g CO2 contains 11.2L volume at STP
22.4L contains 6.022* 10^23 So 11.2L will contain 6.022 * 10^23/2=3.011*10^23 molecules of CO2
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