First you figure out how many grams of sulfur has been used.
2.080g - 1.659g = 0.421g of S used
Now change all of your grams to mols.
1.659g of Cu / 63.546g/mol = 0.02611mol of Cu
0.421g of S / 32.065g/mol = 0.01313 mol of S
From this you can see that S has less so we divide both numbers by the number of moles of S so we can get a ratio of S to Cu.
0.01313/0.01313 = 1 for S
0.02611/0.01313 = 2 for Cu
So the empirical formula would be Cu2S or copper sulfide.
It is 2 because if you have a chart that shown all the elements at the top left corner it would show a number and that is the atomic number
Lead(II) nitrate will react with iron(III) chloride to produce the precipitate lead(II) chloride as shown in the balanced reaction
2FeCl3(aq) + 3Pb(NO3)2(aq) → 2Fe(NO3)3(aq) + 3PbCl2(s)
Calculating the amount of the precipitate lead(II) chloride each reactant will produce:
mol PbCl2 = 0.050L Pb(NO3)2 (0.100mol/1L)(3mol PbCl2/3mol Pb(NO3)2)
= 0.00500mol PbCl2
mol PbCl2 = 0.050L FeCl3 (0.100mol FeCl3/1L)(3mol PbCl2/2mol FeCl3) = 0.00750mol PbCl2
The reactant Pb(NO3)2 produces a lesser amount of the precipitate PbCl2, therefore, the lead(II) nitrate is the limiting reagent for this reaction.
For this case, we use the equation for an ideal gas which is expressed as PV=nRT where P is the pressure, V is the volume, n is the number of moles and T is the temperature. We calculate as follows:
PV = nRT
T = PV / nR
T = 20 kPa (100 L) / 1 mol (8.314)
T = 240.56 K
A standardized test of what you want to do with your friends and you get to know what you think you want to be tre