L have 56 L of H_2H2 gas at STP. Haw many grams do I have of H_2H2 gas?

1 Answer
Dec 15, 2015

The mass of "56 L H"_2"56 L H2 at "STP"STP is "5.0 g"5.0 g.

Explanation:

Use the ideal gas law to determine moles of "H"_2"H2. Then multiply the moles times the molar mass. "STP=273.15 K and 100 kPa"STP=273.15 K and 100 kPa.

Part 1: Ideal Gas Law
PV=nRTPV=nRT, where nn represents moles and RR is the gas constant.

Given/Known
P="100 kPa"P=100 kPa
V="56 L"V=56 L
R="8.3144598 L kPa K"^(-1) "mol"^(-1)"R=8.3144598 L kPa K1mol1
T="273.15 K"T=273.15 K

Unknown
nn

Solution
Rearrange the equation to isolate nn and solve.

n=(PV)/(RT)n=PVRT

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"