If a Eudiometer collects H2 from a reaction with Mg, under the conditions listed below, what's wrong with these calculations?

(Mg + 6M solution of HCl)

Temp of water in eudiometer = 23 C
Room pressure = 103 kPa
Equalized volume of gas in eudiometer = 35.64

I'm trying to find the mass of H2enter image source here

1 Answer
Truong-Son N.
Jun 1, 2018

You're not at STP. STP is #0^@ "C"# and (in most general chemistry textbooks) #"1 atm"#. You are at #23^@ "C"# and about #"1.017 atm"#.

If you have #"35.64 mL"# of #"H"_2(g)#, then you have #"0.03564 L"#. The ideal gas law states:

#PV = nRT#

The volume per mol, #V/n#, is, assuming ideality:

#V/n = (RT)/P#

#= ("0.08206 L"cdotcancel"atm""/mol"cdotcancel"K"cdot(23+273.15) cancel"K")/(103 cancel"kPa" xx cancel"1 atm"/(101.325 cancel"kPa"))#

#=# #"23.91 L/mol H"_2(g)#

As a result, this volume of #"H"_2(g)# contains

#0.03564 cancel"L" xx "1 mol H"_2/(23.91 cancel"L") = "0.00149 mols"#

of #"H"_2(g)#. Therefore, the mass is:

#color(blue)(m_(H_2(g))) = 0.00149 cancel("mols H"_2) xx "2.016 g H"_2/(cancel("1 mol H"_2)) = color(blue)("0.00301 g H"_2(g))#

This is actually over #5%# different from your result, so this is significantly different.

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