Question

What is the mass, in grams, of 6.82 L of nitrogen monoxide at STP? R = 0.0821 L atm / mol K

Answer 1

Answer: 9.13g NO

Step-by-step explanation:

First, I list what I know, which is the values of STP, the volume, the constant R, and the molar mass of NO; I also include what I'm looking for, which is n, the number of moles of NO.

STP = 273.15K, 1 atm

P = 1 atm

V = 6.82L

R = 0.0821 L*atm/mol*K

T = 273.15K

n = ?

NO mass: 30.01 g/mol

Then, I use the ideal gas law, PV = nRT, to find the number of moles of NO:

(1 atm)(6.82 L) = n(
`(0.0821 L*atm)/(mol*K)`)(273.15K)
(1 atm)(6.82 L) = n(
`(22.43 L*atm)/(mol)`)
(1 atm)(6.82 L) / (
`(22.43 L*atm)/(mol)`) = n => divided both sides by 22.4L*atm/mol
n = 0.304 mol NO

Then, I use the molar mass of NO, 30.01 g/mol, to convert from mol NO to g NO:

g NO = 0.304 mol NO(
`(30.01g NO)/(mol NO)`)
= 9.13 g NO

I hope this helps! :)