Answer:
94.74 mol NO₂
General Formulas and Concepts:
Chemistry - Stoichiometry
- Using Dimensional Analysis
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
Explanation:
Step 1: Define
3714 L O₂ at STP
Step 2: Identify Conversions
RxN: 7 mol O₂ = 4 mol NO₂
STP: 22.4 L = 1 mol
Step 3: Stoichiometry
[tex]3714 \ L \ O_2(\frac{1 \ mol \ O_2}{22.4 \ L \ O_2} )(\frac{4 \ mol \ NO_2}{7 \ mol \ O_2} )[/tex] = 94.7449 mol NO₂
Step 4: Check
We are given 4 sig figs. Follow sig fig rules and round.
94.7449 mol NO₂ ≈ 94.74 mol NO₂
