Question

Determine the limiting reactant (lr) and the mass (in g) of nitrogen that can be formed from 50.0 g n2o4 and 45.0 g n2h4. some possibly useful molar masses are as follows: n2o4 = 92.02 g/mol, n2h4 = 32.05 g/mol. n2o4(l) + 2 n2h4(l) → 3 n2(g) + 4 h2o(g) determine the limiting reactant (lr) and the mass (in g) of nitrogen that can be formed from 50.0 g n2o4 and 45.0 g n2h4. some possibly useful molar masses are as follows: n2o4 = 92.02 g/mol, n2h4 = 32.05 g/mol. n2o4(l) + 2 n2h4(l) → 3 n2(g) + 4 h2o(g) lr = n2o4, 105 g n2 formed lr = n2h4, 59.0 g n2 formed lr = n2o4, 45.7 g n2 formed no lr, 45.0 g n2 formed lr = n2h4, 13.3 g n2 formed

Answer 1

                                                   N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)
1) to calculate the limiting reactant you need to pass grams to moles.
 moles is calculated by dividing mass by molar mass

mass of N2O4: 50.0 g 
molar mass of N2O4 = 92.02 g/mol
molar mass of N2H4 = 32.05 g/mol.
mass of N2H4:45.0 g

moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of N2H4

 2)
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react  0,54 moles of N2O4, you need 2x0,54 moles of N2H4 moles
N2H4 needed = 1,08 moles.
You have more that 1,08 moles N2H4, so this means the limiting reagent is not N2H4, it's N2O4. The molecule that has molecules that are left is never the limiting reactant.

3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2

4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
$moles= \frac{grams}{molar mass}$             (molar mass of N2= 28)
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams