N2 H4 (1) + O2 N2 (a) + 2H20()
PLEASE HELP ASAP!
How many grams of hydrazine (N2Ha) are needed to produce 52,4 g
water?
a 31.0
O b
b. 186
c. 32.0
d. 46.5

d. 46.5 g

Explanation:

First, you need to start with a properly balanced equation:

N₂H₄(l) + 3O₂(g) -> 2NO₂(g) + 2H₂O(l)

Then, find the moles of water produced:

2H = 2.01568 amu

O = 15.999 amu

18.01468 amu

52.4 g ÷ 18.01468 amu = 2.9087 moles

The mole ratio in our balanced formula between N₂H₄ : H₂O is 1 : 2, so divide moles of H₂O by 2 to get moles of N₂H₄

2.9087 moles ÷ 2 = 1.45435 moles of N₂H₄

Then, calculate the atomic mass of N₂H₄:

2N = 28.0134 amu

4H = 4.03136 amu

32.04476 amu

Finally, calculate the mass in grams of N₂H₄:

1.45435 moles • 32.04476 amu = 46.604 g

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.