00:03
For a, looking at fe 04 -2 -minus, the oxidation state of iron is plus 6, and the electron configuration of iron is argon 3d2.
00:38
This would be to the electron configuration with an oxidation state of plus six.
00:52
For b, let's calculate the volume of nitrogen gas produced.
00:57
Let's first start with our balanced equation.
01:00
2 -feo4 -2 -minus, plus 2 -h3, 10h plus to produce n2 gas, 2 -3 plus, an 8h2o where has to find the volume of nitrogen gas produced starting from 25 milliliters of 0 .243 molar fe 042 minus and 55 milliliters of 1 .45molar ammonia let's first solve for the moles of n2 produced this is going to be based on the fe 042 minus.
01:57
This is 0 .243 molar times 0 .0250 liters.
02:05
This works out to 0 .006075 moles of f .e.
02:16
F .e .o .42 minus.
02:21
And then we'll multiply this by come down here, one mole stochialmetry here, one mole of n2 to two moles of f -e -o -4 -2 -minus, and this would work out to 3 .0375 times 10 to the negative 3 moles of n2.
02:49
And we'll calculate the moles of n2 based on the ammonia...