Chemistry (stoichiometry)

Question

Heating a 6.862 g sample of an ore containing a metal sulfide in excess oxygen produces 1053 mL of SO2

gas measured at 66 oC and 0.972 atm. Calculate the percentage by mass of sulfur in the ore. (R = 0.0821

L·atm/mol·K)

Answer 1

Ideal Gas Law: PV = nRT

P = pressure = 0.972 atm

V = volume in L = 1053 ml x 1 L/1000 ml = 1.053 L

n = moles of gas = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 66ºC + 273 = 339K

Solving for n we have...

n = PV/RT = (0.972)(1.053) / (0.0821)(339)

n = 0.03677 moles of SO₂ gas

grams of S = 0.03677 mols SO2 x 1 mol S / mol SO2 x 32 g S / mol S = 1.177 g S

%S = 1.177 g / 6.862 g (x100%) = 17.15% S