Math, asked by babydoll84, 8 months ago

how many moles of ethane are required to produce 44g of CO2 after the combustion of ethane?​???​

Answers

Answered by Anonymous
13

Reaction for the combustion of the Ethane is ⇒

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2C₂H₆ + 7O₂ ----------→ 4CO₂ + 6H₂O

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From this reaction, 

∵ 4 mole of the Carbon dioxide is prepared by the 2 mole of the Ethane.

∴ 1 mole of the Carbon dioxide is prepared by the 0.5 mole of the Ethane.

∵ 1 mole of the Carbon dioxide = 44 g. 

∴ Moles of the Ethane required for the combustion is 0.5 moles.

✴✌Hope! it helps uhh ✌✴

Answered by ÚɢʟʏÐᴜᴄᴋʟɪɴɢ1
31

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By writing the combustion equation

2CH3_CH3 +7O2----->4CO2 +6H2O

So basicalythe ratio between moles of ethane and carbon dioxide is 1:2

1 mole of ethane---->2moles of CO2

Masses of one mole of each

12×2+6×1 grams of ethane----->2(12+16×2) g CO2

30 grams of ethane--->88g CO2

X grams of ethane---->44g CO2

X=(30×44)/88= 15 grams

No. Of moles = mass(in grams)/molecular mass of ethane

No of moles =15/(12×2)+6×1 = 0.5 molesm

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gd afternoon

hope it's help❤

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