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.

Respuesta :

Answer:

N2O4 is a limiting reactant

Explanation:

The first thing is to write the balanced chemical reaction

[tex]N_{2} O_{4}+2N_{2} H_{4}--->3N_{2} +4H_{2} O[/tex]

Second find the number of moles of each reactant

n(N204) =[tex]\frac{mass}{molar mass}=\frac{50g}{92.02g/mol} =0.54336 mol[/tex]

n(N2H4)=[tex]\frac{45g}{32.05g/mol}[/tex]=1.4041mol

According to the stoichiometrich ratio, one mol of N2O4 two moles of N2H4 to completely react. So double. It follows therefore that that 0.54336 mol of N2O4 will require 1.0867 mol of N2H4 to completely react. BUt as we have seen here, there is more N2H4 than that, it is in excess.

Therefore

N2O4 is a limiting reactant