Ethylene (C 2 H 4 ), an important industrial organic
chemical, can be prepared by heating hexane (C 6 H 14 )
at 8008C:
C6H14 + C2H4
If the yield of ethylene production is 42.5 percent,
what mass of hexane must be reacted to produce 481 g
of ethylene
First, we need to determine the theoretical yield of ethylene. We can do this by setting up a balanced equation for the reaction:
C6H14 + 3C2H4 → 4C2H4
This shows that for every molecule of hexane, we can produce 4 molecules of ethylene. So, we can set up a proportion to find the theoretical yield:
6 moles of hexane : 4 moles of ethylene
x moles of hexane : 1 mole of ethylene
x = (4/6) = 0.67 moles of hexane
Next, we can calculate the mass of ethylene produced from this amount of hexane:
0.67 moles of hexane x (42.5/100) = 0.285 moles of ethylene
Molar mass of ethylene = 28 g/mol
0.285 moles of ethylene x 28 g/mol = 7.98 g
So, 7.98 g of ethylene is the theoretical yield from 0.67 moles of hexane. To produce 481 g of ethylene, we can set up another proportion:
7.98 g of ethylene : 0.67 moles of hexane
481 g of ethylene : x moles of hexane
x = (0.67/7.98) x 481 g ≈ 40.2 g of hexane
Therefore, approximately 40.2 g of hexane must be reacted to produce 481 g of ethylene with a yield of 42.5%.
Ethylene (C 2 H 4 ), an important industrial organic
chemical, can be prepared by heating hexane (C 6 H 14 )
at 8008C:
C6H14→C2H4 + other products
If the yield of ethylene production is 42.5 percent,
what mass of hexane must be reacted to produce 481 g
of ethylene
Since there is no balanced chemical equation provided, we cannot determine the exact stoichiometry of the reaction. However, we can use the given yield of 42.5% to calculate the mass of hexane required to produce 481 g of ethylene.
Let x be the mass of hexane required. The 42.5% yield means that 42.5% of the hexane is converted into ethylene, and the rest remains as "other products". Therefore:
0.425x = mass of ethylene produced
We want to find the value of x that gives us a mass of ethylene equal to 481 g:
0.425x = 481 g
Solving for x, we have:
x = 1130 g
Therefore, 1130 g of hexane must be reacted to produce 481 g of ethylene with a yield of 42.5%.
To determine the mass of hexane required to produce 481 g of ethylene, we need to find the molar mass of ethylene (C2H4) and the balanced chemical equation for the reaction.
Molar mass of ethylene (C2H4):
(2 * atomic mass of C) + (4 * atomic mass of H)
= (2 * 12.01 g/mol) + (4 * 1.008 g/mol)
= 24.02 g/mol + 4.032 g/mol
= 28.052 g/mol
Now, let's write the balanced chemical equation for the reaction:
C6H14 → C2H4
From the balanced equation, we can see that 1 mole of hexane (C6H14) produces 1 mole of ethylene (C2H4).
The molar mass of hexane (C6H14) can be calculated as follows:
(6 * atomic mass of C) + (14 * atomic mass of H)
= (6 * 12.01 g/mol) + (14 * 1.008 g/mol)
= 72.06 g/mol + 14.112 g/mol
= 86.172 g/mol
Now we can calculate the moles of ethylene produced:
Moles of ethylene = mass of ethylene / molar mass of ethylene
= 481 g / 28.052 g/mol
≈ 17.1486 mol
Since the yield of ethylene production is 42.5%, we can calculate the moles of hexane needed:
Moles of hexane = Moles of ethylene / Yield percentage
= 17.1486 mol / 0.425
= 40.3944 mol
Finally, we can determine the mass of hexane needed:
Mass of hexane = Moles of hexane * molar mass of hexane
= 40.3944 mol * 86.172 g/mol
= 3479.44 g
Therefore, approximately 3,479.44 grams (or 3.48 kg) of hexane must be reacted to produce 481 g of ethylene with a yield of 42.5%.