Question

Consider the reaction: FeS(s) + 2HCl (aq) --> FeCl2(aq) + 2H2S(g). A 4.00g sample of FeS containing nonsulfuide impurities reacted with HCl to give 896mL of H2S at 14C and 782 mmHg.Calculate mass percent purity of the sample. State your final answer using 3 Sig figs

Answer 1

molar mass of FeS = 87.91 gm / mole that mean 1 mole of FeS = 87.91 gm

then 4.00 gm of FeS = 4/87.91 = 0.0455 mol

According to reaction 1 mole FeS produce 2 mole of H₂S then 0.0455 mole of FeS produce 0.0455 2 =  0.0910 mole of H₂S

0.0910 mole is therotical yield ( if sample is 100% pure)

now calculate the actual mole of H₂S produced

Use ideal gas equation for calculation of mole of gas

Ideal gas equation

PV = nRT             where, P = atm pressure= 782 mmHg = 1.02895 atm,

V = volume in Liter = 896 ml = 0.896 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 14⁰ = 273.15+ 14 = 287.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (1.028950.896)/(0.08205287.15) = 0.03913 mole of H₂S

According to reaction 1 mole FeS produce 2 mole of H₂S then to produce 0.03913 mole of H₂S sample of FeS require = 0.03913/2 = 0.019565 mole

molar mass of FeS = 87.91 gm / mole that mean 1 mole of FeS = 87.91 gm

then 0.019565 mole of FeS = 0.019565 87.91 = 1.71995915 gm

Thus sample contain only 1.71995915 gm of FeS

mass percent purity of the sample = 1.71995915 100 /4 = 42.9989 %