Question

Some rocket engines use a mixture of hydrazine (N2H4) and hydrogen peroxide (H2O2) as the propellant system. The RXN is given by the equation: N2H4 + 2H2O2 → N2 + 4H2O a. Which is the limiting reactant in theis reaction when 0.7 mol of hydrazine reacts with 0.500 mol of hydrogen peroxide. b. How much of the excess reactant, in moles, remains unused? c. How much of each product, in moles, is formed?

Answer 1

Answer:

a) H2O2 is the limiting reactant

b) There will remain 0.450 moles of N2H4

c) There will be produced 0.250 moles of N2 and 1 mol of H2O

Explanation:

Step 1: Data given

Number of moles of N2H4 = 0.7 moles

Number of moles H2O2 = 0.500 moles

Molar mass of N2H4 = 32.05 g/mol

Molar mass of H2O2 = 34.01 g/mol

Step 2: The balanced equation

N2H4 + 2H2O2 → N2 + 4H2O

Step 3: Calculate the limiting reactant

For 1 mol of N2H4 we need 2 moles of H2O2 to produce 1 mol of N2 and 4 moles of H2O

H2O2 is the limiting reactant. It will completely be consumed. (0.500 moles).

N2H4 is in excess. There will react 0.500/2 = 0.250 moles of N2H4

There will remain 0.700 - 0.250 moles = 0.450 moles of N2H4

Step 4: Calculate moles of products

For 1 mol of N2H4 we need 2 moles of H2O2 to produce 1 mol of N2 and 4 moles of H2O

For 0.500 moles of H2O2. we'll have 0.250 moles of N2 and 1 mol of H2O

Answer 2

Answer:

Explanation:

(a). Hydrogen peroxide is the limiting reactant, because from the rxn the mole ratio is 1:2, therefore, 0.7mol of N2H4 is supposed to react with 1.4mol of H2O2.

(b). From the mole ratio, 0.5mol of H2O2 will react with 0.25mol of N2H4. Therefore, the unused mol of N2H4 will be (0.7-0.25)mol

=0.45mol

(c). 0.25mol of N2 and 1.0mol of H20 are the products formed based on the mole ratio from the rxn.