In this question, we have the following reaction:
2 H2O2 + N2H4 -> 4 H2O + N2
The reaction is already balanced, the molar ratios are:
2 H2O2 = 1 N2H4
2 H2O2 = 1 N2
1 N2H4 = 1 N2
We have the following mass given in the question and the molar mass:
8.82 grams of H2O2, molar mass = 34.01g/mol
6.81 grams of N2H4, molar mass = 32.04g/mol
N2 molar mass = 28.02g/mol
Now we need to find the limiting reactant, we will start by the H2O2
34.01g = 1 mol
8.82g = x moles
x = 0.26 moles of H2O2 in 8.82 grams
According to the molar ratio, we will need the number of moles of H2O2 divided by 2 in order to find the number of moles of N2H4
0.26/2 = 0.13 moles of N2H4
Now we need to check if we have 0.13 moles of N2H4 or we have more of it:
32.04g = 1 mol
6.81g = x moles
x = 0.212 moles of N2H4, but we only need 0.13 moles, therefore we have an excess of N2H4 and H2O2 is the limiting reactant
Now we need to find the final mass of N2, we will do that by using the number of moles of the limiting reactant, H2O2 0.26 moles, and according to the molar ratio between H2O2 and N2, we need 0.13 moles of N2 being produced, using this number of moles and its molar mass, we can find the final mass:
28.01g = 1 mol
x grams = 0.13 moles
x = 3.64 grams of N2 is produced
Answer = 3.64 grams of N2