C6H14 ---> C2H4 + other products
If the yield of ethylene (C2H4) production is 42.5 percent, what mass of hexane (C6H14) must be reacted to produce 481g of ethylene?
Can someone please help me. I dont understand. I'm more interested in the explanation than in the answer.
2006-08-27 05:24:23 · 4 answers · asked by dingdong 1 in Science & Mathematics ➔ Chemistry
I'd help, but you'd think my answer was wrong.
Thanks for the 2 points!!
2006-08-27 22:45:32 · answer #1 · answered by Mr. G 6 · 0⤊ 0⤋
This looks just like a simple ratio problem
42.5 g ethylene/100g hexane = 481 g ethylene/x g hexane
solve for x
2006-08-27 13:10:02 · answer #2 · answered by Anonymous · 0⤊ 1⤋
well in this reaction, from 1 mole of hexane you get one mole of ethylene and various other products, with a yeald of 42.5 %. which means that if normaly from 1 mole of hexane you theoretically get 1 mole of ethylene, in this case you will get only 42.5% of the theoretical quantiy (that would be 0.425 moles ethylene). The molecular mas of ethylene is 28, meaning that 481g ethylene are 17.18 moles. Now comes the calculus part. 0.425 moles of ethylene are obtained from 1 mole of hexane. Then 17.18 moles of ethylene are obtained from 40.42 moles of hexane. Molar mass of hexane is 86, which means the mass of hexzane is 40.42 * 86 = 3476g hexane
2006-08-27 12:35:29 · answer #3 · answered by andreicnx 3 · 0⤊ 0⤋
1 mole of hexane gives 1 mole of ethene (100%)
86 grms makes 28 grms
But at 42.5% only 42.5/100 x 28 = 11.9 grms
So you could make 1 gr of ethene with 86/11.9 grms hexane
= .607 grms
To make 481 grms you would need 481 x .607
=292.1grms
2006-08-27 14:12:15 · answer #4 · answered by lykovetos 5 · 0⤊ 0⤋