Consider the reaction between N2H4 and N2O4: 2N2H4(g)+N2O4(g)3N2(g)+4H2O(g) A re
ID: 528213 • Letter: C
Question
Consider the reaction between N2H4 and N2O4: 2N2H4(g)+N2O4(g)3N2(g)+4H2O(g) A reaction vessel initially contains 28.5 gN2H4 and 74.9 g of N2O4.
Part A Calculate the mass of N2H4 that will be in the reaction vessel once the reactants have reacted as much as possible. (Assume 100% yield.)
Part B Calculate the mass of N2O4 that will be in the reaction vessel once the reactants have reacted as much as possible. (Assume 100% yield.) Express your answer with the appropriate units.
Part C Calculate the mass of N2 that will be in the reaction vessel once the reactants have reacted as much as possible. (Assume 100% yield.)
Part D Calculate the mass of H2O that will be in the reaction vessel once the reactants have reacted as much as possible. (Assume 100% yield.)
Explanation / Answer
A)
molar mass of N2H4 = 32.05 g/mol
mol of N2H4 = (mass)/(molar mass)
= 28.5/32.05
= 0.89 mol
mass of N2O4 = 74.9 g
molar mass of N2O4 = 92.02 g/mol
mol of N2O4 = (mass)/(molar mass)
= 74.9/92.02
= 0.81 mol
Balanced chemical equation is:
2N2H4 + 1N2O4 ---> 3N2 + 4H2O
2 mol of N2H4 reacts with 1 mol of N2O4
for 0.89 mol of N2H4 0.44 mol of N2O4 is required
But we have 0.81 mol of N2O4
N2H4 is limiting reagent
we will use N2H4 in further calculation
So, mass of N2H4 remaining is 0 g
B)
mol of N2O4 reacted = (1/2)*mol of N2H4
= (1/2)*0.89
= 0.445 mol
mass of N2O4 reacted = number of mol * molar mass of N2O4
= 0.445 mol * 92.02 g/mol
= 40.9 g
mass of N2O4 remaining = 74.9 - 40.9 = 34.0 g
C)
According to balanced equation
mol of N2 formed = (3/2)* moles of N2H4
= (3/2)*0.89
= 1.33 mol
mass of N2 = number of mol * molar mass
= 1.33*28
= 37.35 g
D)
According to balanced equation
mol of H2O formed = (4/2)* moles of N2H4
= (4/2)*0.89
= 1.78 mol
masss of H2O = number of mol * molar mass
= 1.78*18
= 32.0 g