Question

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❥ʜᴏᴡ ᴍᴜᴄʜ ᴄᴀʀʙᴏɴ ᴅɪᴏxɪᴅᴇ ᴡɪʟʟ ʙᴇ ᴏʙᴛᴀɪɴᴇᴅ ᴏɴ ʙᴜʀɴɪɴɢ 1ʟ ᴏғ ᴇᴛʜʏʟᴇɴᴇ (ᴄ2ʜ4) ɪɴ 5ʟ ᴏғ ᴏxʏɢᴇɴ ᴀɴᴅ ʜᴏᴡ ᴍᴜᴄʜ ᴏxʏɢᴇɴ ᴡɪʟʟ ʙᴇ ʟᴇғᴛ ᴜɴᴜsᴇᴅ ?

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Answer 1

C2H4 + 3O2 → 2CO2 + 2H2O

So,

1 mole of C2H4 reacts with 3 moles of O2 in order to form 2 moles of CO2.

1mole = 22.4l

So, 22.4l of C2H4 reacts with (3×22.4)l of O2 in order to form (2 ×22.4)l of CO2.

We have, 1l of C2H4 and 5l of O2.

We observed that,

22.4l of C2H4 reacts with (3×22.4)l of O2

So, 1l of C2H4 will react with 3×22.4/22.4 = 3l of O2

So, 2l of O2 is left behind.

C2H4 is available is limited amount, so it's the limiting reagent. And stoichiometric calculation are done with respect to the limiting reagent (if available).

Above we talked about,

22.4l of C2H4 reacts with (3×22.4)l of O2 in order to form (2 ×22.4)l of CO2.

Since, C2H4 is limiting reagent, O2 is not considered in the calculations.

So, 22.4l of C2H4 reacts with O2 in order to form (2 ×22.4)l of CO2.

Thus, 1l of C2H4 will react with O2 to form 2×22.4/22.4 = 2l of CO2.

Therefore, 2l of CO2 is formed.

Thanks. :)

Answer 2

Answer:

ki s hu un ni e w he re r ya.. TwT

Explanation:

Mi ss u s o mu c h....c o me ba c k pweasee...