00:01
In stochometry, we relate the quantities of the substances that are involved in a chemical reaction.
00:06
In this problem, we're given with a balanced chemical equation for the reaction between so2 and oxygen gas to form sulfur dioxide.
00:16
Now, if we start with 285 .5m.
00:21
Of so2 and 158 .9m.
00:28
Of o2, we wish to find, the following so by the way these gases are measured at stp that means standard temperature and pressure so for the first one we want to know the limiting reactant it's called that as lr and for the next one we're going to find the theoretical yield of so3 that is the amount of so3 that is formed from the reaction and for the next one we're going to solve for the percent yield that's b -y if we obtain if the actual amount of so3 gas obtained at stp is 2 .805 grams at stpa.
01:15
So let's start by establishing the mole relationship from the balance equation.
01:22
We see that for every two moles of so3, or rather so2, one mole of o2 react to form two moles of so3.
01:33
So under the same conditions such as stp, this small relationship is the same as volume relationship so that 2 liters of s2 will react with 1 liter of o2 to form 2 liters of so3.
01:48
So of course this can also mean 2ml, 1ml and 2ml of the gases.
01:55
So let us solve for the ml of so3 formed from the given, from each of the given ml.
02:03
Of the reacted gases so such as 285 .5 m l of so2 so for that we will simply use the volume relationship that we have established a while ago where for every two moles of or 2 ml of so2 we form 2ml of so3 also so this gives us 285 .m .m...