You burn 6 grams of carbon in plenty of air: C ( s ) + O 2 ( g ) → CO 2 ( g ) What volume of gas will form at rtp?

C ( s ) + O 2 ( g ) → CO 2 ( g ) From the reaction, it is clear that one mole carbon burns with one mole of oxygen to give one mole of carbon dioxide. Number of moles of 6 grams of carbon is 6 g 12 g = 0 . 5 mol Therefore 0 . 5 mol of carbon will burn with 0 . 5 mol of oxygen to give 0 . 5 mol of carbon dioxide. Since, the molar volume of any gas at rtp is 24 dm 3 . Therefore, the volume of 0 . 5 mol of carbon dioxide is 0 . 5 mol × 24 dm 3 = 12 dm 3

You burn 6 grams of carbon in plenty of air: C ( s ) + O 2 ( g ) → CO 2 ( g ) What volume of oxygen at rtp will be used up?

C ( s ) + O 2 ( g ) → CO 2 ( g ) From the reaction, it is clear that one mole carbon burns with one mole of oxygen to give one mole of carbon dioxide. Number of moles of 6 grams of carbon is 6 g 12 g = 0 . 5 mol Therefore 0 . 5 mol of carbon will burn with 0 . 5 mol of oxygen. Since, the molar volume of any gas at rtp is 24 dm 3 . Therefore, the volume of 0 . 5 mol of oxygen gas is 0 . 5 mol × 24 dm 3 = 12 dm 3

If You burn 6 grams of carbon in limited air the reaction is as follows: 2 C ( s ) + O 2 ( g ) → 2 CO ( g ) What volume of gas will form at rtp?

2 C ( s ) + O 2 ( g ) → 2 CO ( g ) From the reaction, it is clear that two moles of carbon burns with one mole of oxygen to give two moles of carbon monoxide. Number of moles of 6 grams of carbon is 6 g 12 g = 0 . 5 mol Therefore 0 . 5 mol of carbon will burn with 0 . 25 mol of oxygen to produce 0 . 5 mol of carbon monoxide. Since, the molar volume of any gas at rtp is 24 dm 3 . Therefore, the volume of 0 . 5 mol of carbon monoxide gas is 0 . 5 mol × 24 dm 3 = 12 dm 3