consider the following balanced equation. N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g) Determine the theoretical yield in gram of nitrogen, N2 , that can be formed from 60.0 g N2O4 and 55.0 g N2H4 ? If the actual experimental yield for N2 is 37.2 g, what is the percent yield of N2? Some possibly useful molar masses are as follows: N2O4 = 92.02 g/mol, N2H4 = 32.05 g/mol, N2 = 28.00 g/mol
consider the following balanced equation.
N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)
Determine the theoretical yield in gram of nitrogen, N2 , that can be formed from 60.0 g N2O4 and 55.0 g N2H4 ?
If the actual experimental yield for N2 is 37.2 g, what is the percent yield of N2?
Some possibly useful molar masses are as follows: N2O4 = 92.02 g/mol, N2H4 = 32.05 g/mol, N2 = 28.00 g/mol
(Hint: the theoretical yield is determined from the limiting reactant. So, you have to find how much gram of N2 is produced from 50.0 g of N2O2 and how much gram of N2 is produced from 45.0 g of N2H4. The one, either the N2O4 or N2H4 , produces the least gram of N2 is called the limiting reactant and the amount of N2 produced from the limiting reactant is called the theoretical yield. Finally, % yield = (Actual / theoretical) x 100 =
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