Question

Volume at NTP of oxygen required to completely
burn 1 kg of coal (100% carbon)
(1) 22400 L
(2) 22.4 * 103 L
(3) 1.86 x 103 L
(4) 1000 L​

Answer 1

Answer:

equation: C + O2 → CO2 1mol C will react with 1 mol O2 Mass ratio 12:32 1kg coal will require 1*32/12 = 2.667kg = 2,667 g O2 Molar mass O2 = 32g/mol 2,667g = 2667/32 = 83.34mol Ar NTP 1 mol = 24L 83.34 mol = 83.34*24 = 2000L You ask about NTP = my understanding is that at NTP 1 mol gas = 24L But at STP , 1 mol gas = 22.4L At STP , 83.34mol = 83.34*22.4 = 1,867L OR 1.86*10^3 L Volume of O2 required art STP = 1.86*10^3 L

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Answer 2

Answer:

Mass of C = 1000 g

1 moles C requires 1 mole O₂    to burn.

Thus, 83.34 mole C requires 83.34 moles O₂  

At NTP, 1 mole gas occupies 22.4 litres.

83.34 moles occupies 83.34 × 22.4 = 1.86 × 10³  L

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Explanation: