00:01
In this problem, we are given that there is a reaction between ethylene gas with water to yield ethyl alcohol.
00:11
This is our balanced chemical equation.
00:28
In part a of the problem, we are given that there are 0 .13 moles of h2o.
00:43
We are trying to solve how many grams of ethylene are needed to react with 0 .133 moles of h2o and how many grams of ethyl alcohol will result.
00:55
Looking at this balanced equation, since everything is in the 1 -1 -to -1 ratio, 0 .13 moles of h2o would indicate that there would be 0 .13 moles of h2o would be 0 .13 moles.
01:12
Of ethylene gas as well.
01:20
We can use this relationship.
01:22
We're trying to find how many grams of ethylene gas is involved in the reaction.
01:32
And this is our molar mass for ethylene gas.
01:37
We are trying to solve for the moles, which we know will be 0 .133 moles of ethylene gas.
01:55
Now we can solve for our x grams, of ethylene gas.
02:00
By solving this in our calculator, we will obtain 3 .72 grams of c2h4.
02:16
The second part, we will have to solve how many grams of ethyl alcohol will result.
02:22
We now know that we have 3 .72 grams of ethylene gas...