00:01
Mtbe which is an additive in gasoline has the structure o -c -h -3 this can be synthesized by addition of methanol across the double bond of two metal propylene this addition is done in presence of water molecule acid in presence of h -plus methanol is added across the double bond to form the ether now the requirement of 2 % of 2 .7 % of oxygen in the gasoline is to be satisfied by addition of mtbe so 2 .7 % 2 .7 grams of oxygen should come from should be from mtbe moral mass of mtbe is 88 .15.
02:10
So it is 88 .15 is a molar mass.
02:17
Now 16 grams of oxygen is obtained from 88 .15 gram of mtbe.
02:41
So 2 .7 grams of oxygen can be obtained from 2 .7 grams of oxygen can be obtained from 2 .3.
03:04
Multiplied by 88 .15 by 16.
03:22
This corresponds to 14 .87 grams.
03:36
So the mixture if the mixture is 100 gram it will contain it should contain 14 .87 grams of mtv and 100 minus 14 .7 that is 85 .12 grams of gasoline that is 85 .12 grams requires 14 .87 grams of mtp therefore 100 grams of gasoline will require 100 grams of gasoline needs 14 .87 into 100 divided by 8785 .12 equals 17 .47 .47 grams of mtbe.
05:29
So this is the first part of the problem.
05:36
Now in the second part, the second part, now the third part, how many details of mtbe would be in each little of fuel mixture? to solve this, let us see what is a mass of one liter of the mixture.
06:02
Density of the mixture equals 0 .740 gram per c...