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What is the volume of 1.51x10^24 molecules of ammonia gas, NH3 at STP?
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1. The first thing you need to do is convert from molecules to moles. \[\frac{ 1molNH3 }{ 6.022x10^{23} } = \frac{ x }{ 1.51x10^{24} }\] x = 2.51 moles NH3 2. One mole of an ideal gas will occupy a volume of 22.4 liters at STP. If you know one mole of NH3 takes up 22.4 liters, you can find how much 2.51 moles will take up. \[\frac{ 1mol }{ 22.4L } = \frac{ 2.51mol }{ x }\] x= (2.51)(22.4) = 56.2L Answer: 56.2 liters NH3
I would use the equationi Pv=nRT
thanks guys!
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