00:01
Using the information given, let's calculate the weight percent of nh4 -2 -s -o -4, ammonium sulfate, in the 0 .47 gram sample.
00:10
Let's write down our equations here.
00:12
Let's start with nh4 to s -o -4, balanced equation, 2 -k -o -h, yields 2, nh3, k2s -4, and 2h -20.
00:30
We also have nh3 plus excess hcl to produce nh4 cl.
00:45
And we have the hcl that's remaining plus n .a .o .h to produce nacl and h2o.
01:02
So first of all, we're going to calculate the moles of the remaining amount of hcl from n -a -o -h.
01:15
So h -c -l remaining is going to be from the titration of n -a -o -h, 0 .121 -molar noh times the volume .0 -1 -1 -1 liters times 1 mole of n .a .o .h.
01:40
To 1 mole of hcl.
01:43
And this will work out to 1 .34.
01:47
We'll keep some extra sig figs here.
01:49
3 .1 times 10 to the negative 3 moles of hcl.
01:52
So that's the amount of hcl remaining...