Find out the volume of 6.023*10²² of ammonia at stp?

1 Answer
Aug 17, 2017

The volume is 2.271 L.

Explanation:

Method 1. Using the Ideal Gas Law

We can use the Ideal Gas Law to solve this problem.

color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "

where

  • p is the pressure
  • V is the volume
  • n is the number of moles
  • R is the gas constant
  • T is the temperature

We can rearrange the Ideal Gas Law to get

V = (nRT)/p

Step 1. Calculate the moles of ammonia

n = 6.023 × 10^22 color(red)(cancel(color(black)("molecules NH"_3))) × "1 mol NH"_3/(6.022 × 10^23 color(red)(cancel(color(black)("molecules NH"_3)))) = "0.100 02 mol NH"_3

Step 2. Calculate the volume at STP

Remember that STP is defined as 0 °C and 1 bar.

n = "0.100 02 mol"
R = "0.083 14 bar·L·K"^"-1""mol"^"-1"
T = "(0 + 273.15) K" = "273.15 K"
p = "1 bar"

V = (nRT)/p = ("0.100 02" color(red)(cancel(color(black)("g"))) × "0.083 14" color(red)(cancel(color(black)("bar"))) ·"L"· color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = "2.271 L"

Method 2. Using the molar volume

We know that there are 0.100 02 mol of "NH"_3.

We also know that the molar volume of a gas is 22.71 L at STP.

V = "0.100 02" color(red)(cancel(color(black)("mol NH"_3))) × "22.71 L NH"_3/(1 color(red)(cancel(color(black)("mol NH"_3)))) = "2.271 L NH"_3