Consider the reaction: 2 N_2 H_4 (g) + N_2Q_4(g) implies 3 N_2(g) + 4 H_2 O (I)

Question

Consider the reaction: 2 N_2 H_4 (g) + N_2Q_4(g) implies 3 N_2(g) + 4 H_2 O (I) If 18.2 g of N_2 H_4 are mixed with 23.0 g of N_2 O_4 (g), What is the limiting reagent? Show all work. A guess is not sufficient How many grams of H_2 O be produced under these conditions? If 0.446 g of H_2 O obtained are in lab when the reaction is run under these conditions, What is the percent yield? Molar mass of N_2 H_4 = 32.05 Molar mass of N_2 = 28.02 Molar mass of N_2 O_4 = 92.02 Molar mass of H_2 O (I) - 18.02

Explanation / Answer

Given,

Mass of N2H4 = 18.2 g

Mass of N2O4 = 23 g

=> Moles of N2H4 = 18.2 / 32.05 = 0.568 moles

and Moles of N2O4 = 23 / 92.02 = 0.25 moles

The reaction is,

2N2H4 + N2O4 ---> 3N2 + 4H2O

According to the stoichiometry of the reaction 2 moles of N2H4 reacts with 1 mole of N2O4 to produce 3 moles of N2 and 4 moles of H2O

Since, 1 mole N2O4 reacts with 2 moles N2H4

0.25 moles of N2O4 will react with 0.5 moles of N2H4 to produce 1 mole of H2O

After this N2O4 will completely exhaust and N2H4 will be remaining

i) N2O4 is the Limiting reagent and N2H4 is in excess

ii) Moles of H2O produced = 1 mole

=> Mass of H2O produced = 1 x 18.02 = 18.02 g

iii) % yield = (0.446 x 100) / 18.02 = 2.475 %

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---