Question

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

Answer 1

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`