Answer:
See explanation below
Explanation:
I found a picture of this exercise, to show you how to do this problem.
The first picture is the compound, and the second is the mechanism of reaction to do this claisen rearrangement.
Hope this helps
Your skin is an important organ and has several functions. All of the functions below are performed by the skin EXCEPT
1 answer:
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Answer:
a) molar composition of this gas on both a wet and a dry basis are
5.76 moles and 5.20 moles respectively.
Ratio of moles of water to the moles of dry gas =0.108 moles
b) Total air required = 68.51 kmoles/h
So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.
Explanation:
Let assume we have 100 g of mixture of gas:
Given that :
Mass of methane =75 g
Mass of ethane = 10 g
Mass of ethylene = 5 g
∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)
Their molar composition can be calculated as follows:
Molar mass of methane
Molar mass of ethane
Molar mass of ethylene
Molar mass of water
number of moles =
Their molar composition can be calculated as follows:
4.69 moles
0.33 moles
0.18 moles
0.56 moles
Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles
= 5.76 moles
Total moles of gas for dry basis = (5.76 - 0.56)moles
= 5.20 moles
Ratio of moles of water to the moles of dry gas =
=
= 0.108 moles
b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:
4.69 2× 4.69
moles moles
0.33 3.5 × 0.33
moles moles
0.18 3× 0.18
moles moles
Mass flow rate = 100 kg/h
Their Molar Flow rate is as follows;
Total moles of required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles
= 11.075 k moles.
In 1 mole air = 0.21 moles
Thus, moles of air required =
= 52.7 k mole
30% excess air = 0.3 × 52.7 k moles
= 15.81 k moles
Total air required = (52.7 + 15.81 ) k moles/h
= 68.51 k moles/h
So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.