00:01
We're told that zinc metal reacts with silver nitrate to produce silver metal and aqueous zinc nitrate solution.
00:07
We're given the equation of the reaction.
00:10
It's a single replacement equation of hno3 and we have znno3 bracket 2 plus ag.
00:25
Because we're going to have to balance this this is balanced now.
00:31
Now we're told what mass of silver will be produced when 25 grams of zinc is added to a beaker containing 105 .5 grams of zinc nitrate dissolved in 250 mls of water.
00:50
So what this means is that first we can find the concentration of the zinc nitrate, find the moles of that and find the number of moles of the zinc we're given.
01:00
So we have to first determine which of them is the limiting reagent.
01:06
So what i'm going to say, what we have to do is find the number of moles of the zinc that was given to us would have been 25 divided by the relative atomic mass of zinc which is 25 .00 which is 65 .38 and when you divide this you have 0 .38 moles.
01:37
Well the number of moles of the silver nitrate that we were provided with would have been the mass which we were given as 105 .5 grams divided by the molar mass of silver nitrate is 169 .87.
02:00
And when you divide this you would have 0 .62 moles.
02:06
So these are the number of moles of the reactant.
02:11
Now we have to determine one of them is the limiting reactant.
02:14
Only one of them is completely used up.
02:17
And the only way we can deduce that is to try to find the number of moles of silver that is produced from each of them...