Answer:
63.5 w isvthebanswerok is th answer
Valence electrons determine its valency ? (The group, as in its oxidation state and stability)
1. The balanced equation tells us that 2 moles of H2S produce 2 moles of H2O.
8.3 moles H2S x (2 moles H2O / 2 moles H2S) = 8.3 moles H2O = theoretical amount produced
8.3 moles H2O x (18.0 g H2O / 1 mole H2O) = 149 g H2O produced theoretically
% yield = (actual amount produced / theoretical amount) x 100 = (137.1 g / 149 g) x 100 = 91.8 % yield
2. Calculate moles of each reactant.
150.0 g N2 x (1 mole N2 / 28.0 g N2) = 5.36 moles N2
32.1 g H2 x (1 mole H2 / 2.02 g H2) = 15.9 moles H2
The balanced equation tells us that we need 3 moles of H2 to react with every 1 mole of N2.
So if we have 5.36 moles N2, we need 3x that = 16.1 moles H2. Do we have that much available? No, just under at 15.9 moles. So H2 is the limiting reactant. At the end of the reaction there will be a little N2 left over.
Answer:
Theoretical yield of HI is 512 g.
The percent yield for this reaction is 25%.
Explanation:

Moles of hydrogen gas = 3.0 moles
Moles of iodine gas = 2.0 moles
According to reaction 1 mol of hydrogen gas reacts with 1 mol of iodine gas.
Then 3.0 moles of hydrogen gas reacts with 3.0 mol of iodine gas. But there are 2.0 moles of iodine gas. Hence,Iodine is a limiting reagent. The production of HI will depend upon iodine gas moles.
According to reaction , 1 mol of iodine gas gives 2 moles of HI.
Then 2 moles of iodine gas will give:
of HI
Theoretically we will get 4 moles of HI.
Theoretical yield of HI = 4 mol × 128 g/mol= 512 g
Experimental yield of HI = 1.0 mol
= 1 mol × 128 g/mol= 128 g


The percent yield for this reaction is 25%.