What is the number of moles in 9.63 L of #H_2S# gas at STP?

2 Answers
May 19, 2016

Answer:

0.430 mol (rounded with significant figures)

Explanation:

Any gas, at STP, will have 22.4 L per mol. Taking this into account, using dimensional analysis:

#9.63 L H_2S * (1 mol H_2S)/(22.4 L H_2S)#

#(9.63 L H_2S)/(22.4 L H_2S)=0.4299 mol#

Which rounds to 0.430 mol, with three significant figures

May 19, 2016

Answer:

#P.V=n.R.T#

Explanation:

Use the ideal gas equation

#P.V=n.R.T#

Solve for #n#,

#n=( P*V)/(R.T)#

#P" is the pressure of the gas (" atm")"#
#V" is the volume of the gas ("L")"#
#T" is the Kelvin temperature ("K")"#
#R" is the universal gas constant ("0.0821L.\atm.mol^-1.K^-1")" #

#n=(1.00*atm*9.63*L) /(0.0821*L*atm*mol.^-1*K^-1*273*K#

#n=(1.00*cancel(atm)*9.63*cancel(L)) /(0.0821*cancel(L)* cancel(atm)*mol.^-1*cancel(K^-1)*273*cancel(K)#

#n=0.430* mol.#