Question #12c2c

1 Answer
Jul 8, 2017

Answer:

#"26 g"#

Explanation:

The idea here is that you're looking for the mass of the metal that will displace exactly #1.008# parts by mass of elemental hydrogen, #"H"#.

The first thing that you need to do here is to use the ideal gas law equation

#color(blue)(ul(color(black)(PV = nRT)))#

Here

  • #P# is the pressure of the gas
  • #V# is the volume it occupies
  • #n# is the number of moles of gas present in the sample
  • #R# is the universal gas constant, equal to #0.0821("atm L")/("mol K")#
  • #T# is the absolute temperature of the gas

to find the number of moles of hydrogen gas, #"H"_2#, present in your sample under normal conditions for pressure and temperature, i.e. at NTP, which are defined as a pressure of #"1 atm"# and a temperature of #20^@"C = 293.15 K"#,

Rearrange the equation to solve for #n#

#PV = nRT implies n = (PV)/(RT)#

Plug in your values to find

#n = (1 color(red)(cancel(color(black)("atm"))) * 1.12color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 293.15 color(red)(cancel(color(black)("K")))) = "0.04654 moles H"_2#

The number of moles of elemental hydrogen present in your sample will be equal to

#0.04654 color(red)(cancel(color(black)("moles H"_2))) * "2 moles H"/(1color(red)(cancel(color(black)("mole H"_2)))) ~~ "0.09308 moles H"#

Convert this to grams by using the molar mass of elemental hydrogen

#0.09308 color(red)(cancel(color(black)("moles H"))) * "1.008 g"/(1color(red)(cancel(color(black)("mole H")))) = "0.09382 g"#

Now, if #"0.09382 g"# of hydrogen were displaced by #"2.4 g"# of metal, it follows that #"1.008 g"# of hydrogen will be displaced by

#1.008 color(red)(cancel(color(black)("g H"))) * "2.4 g metal"/(0.09382color(red)(cancel(color(black)("g H")))) = color(darkgreen)(ul(color(black)("26 g metal")))#

The answer is rounded to two sig figs, the number of significant figures you have for the mass of the metal.

#color(white)(a)#
SIDE NOTE I suspect that the problem was designed with the old STP conditions in mind, i.e. a pressure of #"1 atm"# and a temperature of #0^@"C"#.

Under these conditions for pressure and temperature, the molar volume of an ideal gas is equal to #"22.4 L"#.

In this case, the number of moles of elemental hydrogen would come out to be equal to #0.100#, which would produce an equivalent mass of #"24 g"# for the metal.