Consider the reaction between N2H4 and N2O4: 2N2H4(g)+N2O4(g)3N2(g)+4H2O(g) A re

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

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