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

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Aug 17, 2017

Answer:

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#

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