What is the number of moles in 9.63 L of H_2SH2S gas at STP?

2 Answers
May 19, 2016

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)9.63LH2S1molH2S22.4LH2S

(9.63 L H_2S)/(22.4 L H_2S)=0.4299 mol9.63LH2S22.4LH2S=0.4299mol

Which rounds to 0.430 mol, with three significant figures

May 19, 2016

P.V=n.R.TP.V=n.R.T

Explanation:

Use the ideal gas equation

P.V=n.R.TP.V=n.R.T

Solve for nn,

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

P" is the pressure of the gas (" atm")"P is the pressure of the gas (atm)
V" is the volume of the gas ("L")"V is the volume of the gas (L)
T" is the Kelvin temperature ("K")"T is the Kelvin temperature (K)
R" is the universal gas constant ("0.0821L.\atm.mol^-1.K^-1")" R is the universal gas constant (0.0821L.atm.mol1.K1)

n=(1.00*atm*9.63*L) /(0.0821*L*atm*mol.^-1*K^-1*273*Kn=1.00atm9.63L0.0821Latmmol.1K1273K

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.