00:01
So to solve for this, let us write first d equation for this reaction.
00:06
We have silver nitrate, that's a -g -n -o -3.
00:10
This is silver -nitrate, and then barium chloride, that's b -a -c -l -2.
00:15
This will produce silver chloride, that's a -g -c -l, and then you have barium nitrate that's b, a -n -o -3, and two.
00:25
Okay, so to balance this, we need to write here two, or...
00:30
We have two chlorines on the left side of the equation so that means we need to write here two also okay so that's the balance equation now and our our unknown here is a percent yield percent yield is equivalent to the actual yield this is actual yield divided by the theoretical yield multiplied by a hundred percent okay so we have here to given the actual so that means we need to calculate first the theoretical yield.
01:05
And you could do that using stochometry.
01:08
Okay, so theoretical yield, the amount of silver chloride produced, egcl.
01:15
If we have 5 .95 grams of ag no3, the silver nitrate, convert this to most by dividing the molar mass of silver nitrate, that's 169 .87 grams per mole.
01:31
Okay and then as per the balance equation for every two moles of agn03 so produce two moles of agcl too also two moles of agcl rather silver chloride okay convert this to mass by multiplying the molar mass of silver chloride and that's 140 3 .32 grams okay divided by moles of agcl okay, so you can see you'll be able to cancel this...