L have 56 L of #H_2# gas at STP. Haw many grams do I have of #H_2# gas?

1 Answer
Dec 15, 2015

Answer:

The mass of #"56 L H"_2"# at #"STP"# is #"5.0 g"#.

Explanation:

Use the ideal gas law to determine moles of #"H"_2"#. Then multiply the moles times the molar mass. #"STP=273.15 K and 100 kPa"#.

Part 1: Ideal Gas Law
#PV=nRT#, where #n# represents moles and #R# is the gas constant.

Given/Known
#P="100 kPa"#
#V="56 L"#
#R="8.3144598 L kPa K"^(-1) "mol"^(-1)"#
#T="273.15 K"#

Unknown
#n#

Solution
Rearrange the equation to isolate #n# and solve.

#n=(PV)/(RT)#

#n=(100cancel"kPa"xx56cancel"L")/(8.3144598cancel"L" cancel"kPa" cancel"K"^(-1) "mol"^(-1)xx273.15cancel"K")="2.5 mol H"_2"# (rounded to two significant figures)

Part 2: Mass of #"H"_2"#
Multiply the moles of hydrogen gas times its molar mass, #"2.01588 g/mol"#.

#2.5cancel"mol H"_2xx(2.01588"g H"_2)/(1cancel"mol H"_2)="5.0 g H"_2"#