84 J/6 moles = 14 J/mole = ∆H
-14 J/mol
-14 J/Mol
84 J/6 moles = 14 J/mole = ∆H
-14 J/mol
This is not a common reaction at standard temperature and pressure.
N2 + 3H2 -> 2NH3 If you have moles produced you can do it this way. 22.5 moles NH3 (3 moles H2/2 moles NH3) = 33.8 moles H2 needed -----------------------------------
-14 J/Mol
84 J/6 moles = 14 J/mole = ∆H
12 J/3 moles = 4 J/mole. Thus, H of reaction in kJ/mole = 0.004 kJ/mole
-14 J/mol
-14 J/mol
This is not a common reaction at standard temperature and pressure.
Carbon dioxide is the limiting reagent.
This is the number before a chemical compound.
The reaction of carbon dioxide and potassium oxide is 4KO2 + 2CO2 = 2K2CO3 + 3O2. 156 grams of CO2 is 3.54 moles, which will produce 5.31 moles of O2.
N2 + 3H2 -> 2NH3 If you have moles produced you can do it this way. 22.5 moles NH3 (3 moles H2/2 moles NH3) = 33.8 moles H2 needed -----------------------------------
If the reaction were 2AlCl3 + 3Pb ---> 3PbCl2 + 2Al, then from this balanced equation, 14 moles of AlCl3 would produce 3/2 x 14 moles of PbCl2 = 21 moles of PbCl2.
Assuming a decomposition reaction with this equation: 2KClO3(s) --> 2KCl(s) + 3O2(g), the ratio is 2:3, and if you produce 15mol O2, then 10mol potassium chlorate are needed.