<span>4FeS2 + 11O2 = 2Fe2O3 + 8SO2</span>
Percent yield is calculated as the actual yield divided by the theoretical yield multiplied by 100.
Actual yield = 55 g ( 1 mol / 159.69 g ) = 0.34 mol Fe2O3
To find for the theoretical yield, we first determine the limiting reactant.
100 g O2 ( 1 mol / 32 g) = 3.13 mol O2
200 g FeS2 (1 mol / 119.98g) = 1.67 mol FeS2
Therefore, the limiting reactant is O2.
Theoretical yield = 3.13 mol O2 ( 2 mol Fe2O3 / 11 mol O2 ) = 0.57 mol Fe2O3
Percent yield = (0.34 mol / 0.57 mol) x 100 = 59.74%
The answer is to test a piece of untreated and treated under same conditions!
<span>PbO
Let's look at each of the 4 compounds and see what's needed.
PbO.
* Oxygen has a valance shell that's missing 2 electrons and wants to get those 2 elections. Lead donates them, so you have a Lead (II) ions. This is a correct choice.
PbCl4
* Chlorine wants to grab 1 electron to fill it's valance shell and Lead donates that election. However, there's 4 chlorine atoms and every one of them wants and electron, and lead is donating all 4 of the desired electrons making the Lead (IV) ion. So this is a bad choice.
Pb2O
* Oxygen still wants 2 electrons and gets them from the lead. But there's 2 lead atoms and each of them donates 1 election making for 2 Lead(I) ions. So this too is a bad choice.
Pb2S
* Sulfur is in the same column of the periodic table as oxygen and if this compound were to exist would have similar properties as Pb2O and would have Lead(I) ions. So this is a bad choice.</span>
PH = 0.1289<span> for </span>1.50<span> M solution of weak acid with Ka value of </span><span>.73</span>
The gram-formula mass of Sm is 150.36 u, and the gram-formula mass of O is 15.999 u, so the gram-formula mass of SmO is about 150.36+15.999 = 160.36 g/mol.
So, there are about (9.30 * 10^-3)(160.36)=1.49 grams
