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Question

Methane burns in oxygen to form CO2 and H2O. Write a balanced equation for the reaction and calculate: (i) How much amount of CO2 and H2O are formed when 8 g of methane reacts with oxygen? (ii) Which is the limiting reagent when the above amount of methane reacts with 29 g of O2?

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Solution

16g of methane reacMethane burns in oxygen to form carbon dioxide and water.
CH4+2O2 CO2+2H2O
CH4CH_4(i) Molar mass of CH4CH4CH_4
is 12+1×1\times1×4=16g
Molar mass of CO2 is 12+16×\times×2=44g
Molar mass of H2OH_2OH2O is 1×\times×2+16 = 18g
Molar mass of O2O_2O2 is 2×\times×16 = 32g
According to the balanced chemical equation,
1 mole of methane reacts with 2 mole of oxygen to form 1 mole of carbon dioxide and 2 moles of water.
no. of moles = givenmassmolarmass
no. of moles of methane =816 =0.5
So 0.5 moles of methane give 0.5 moles of carbon dioxide and 1 mole of water
Amount of methane = 0.5×16 = 8g
Amount of carbon dioxide = 0.5×44 = 22g
Amount of water = 18×1 = 18g
(ii) As per the equation, 16g of methane reacts with 64g of oxygen to form 44g of carbon dioxide and 32g of water.
Hence 8g of methane reacts with 32g of oxygen only. But the given amount of oxygen is 29g. So oxygen acts as a limiting reagent.


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