00:01
To determine which is the limiting reactant, we need to know how much product each reactant can produce.
00:07
Because the second question deals with determining the mass of lead -2 iodide will calculate the mass of lead -2 iodide that can be produced from each amount of reactant.
00:18
The balanced chemical reaction will require a two in front of potassium nitrate, so we have two nitrates.
00:24
That gave us two potassium, so we need to put a two in front of potassium iodide.
00:29
That'll give us two iodides and we have two iodides.
00:34
So starting with 16 .4 grams of lead to nitrate, we can convert it into moles by dividing by the molar mass led to nitrate and then go from moles led to nitrate to moles led to iodide, seeing it's a one -to -one mole relationship.
00:50
Once we have moles led to iodide, we simply multiply it by the molar mass of lead to iodide to get grams, and 64, i'm sorry, 16 .4, grams lead to nitrate will produce 22 .8 grams led to iodide.
01:08
We then carry out a similar calculation for potassium iodide starting at 28 .5 grams.
01:15
We divide by the molar mass potassium iodide, then convert the moles potassium iodide into moles of lead to iodide with the 2 to 1 mole relationship...