Question

Hydrazine reacts with oxygen according to the following equation: N2H4(g) +O2(g) → N2(g) + 2 H2O(l) How many L of N2, measured at 34.9 °C and 755.08 torr, will be produced at the same time that 914.894 g of H2O is produced?

Answer 1

V ≈ 646.50 L

General Formulas and Concepts:

Chemistry - Gas Laws

• Reading a Periodic Table
• Stoichiometry
• Combined Gas Law: PV = nRT
• R constant - 62.4 (L · torr)/(mol · K)
• Kelvin Conversion: K = °C + 273.15

Step-by-step explanation:

Step 1: Define

RxN: N₂H₄ (g) + O₂ (g) → N₂ (g) + 2H₂O (l)

Given: 34.9 °C, 755.08 torr, 914.894 g H₂O

Step 2: Identify Conversions

Kelvin Conversion

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Step 3: Convert

Stoichiometry:
`914.894 \ g \ H_2O((1 \ mol \ H_2O)/(18.02 \ g \ H_2O) )((1 \ mol \ N_2)/(2 \ mol \ H_2O) )` = 25.3955 mol N₂

Temperature: 34.9 + 273.15 = 308.05 K

Step 4: Find Volume

1. Substitute variables: (755.08 torr)V = (25.3955 mol)(62.4 (L · torr)/(mol · K))(308.05 K)
2. Multiply: (755.08 torr)V = 488160 L · torr
3. Isolate V: V = 646.502 L

Step 5: Check

We are given 5 sig figs as our lowest. Follow sig fig rules and round.

646.502 L ≈ 646.50 L