Percent by mass of sulfur in ore?

[匿名]

2016-10-18 2:28 am

Heating a 6.599 g sample of an ore containing a metal sulfide, in the presence of excess oxygen, produces 1.462 L of dry SO2, measured at 47.4 °C and 753.5 torr. Calculate the percent by mass of sulfur in the ore. Assume that all sulfur in the sample was converted to SO2.
S(s) + O 2 (g) → SO 2 (g)

回答 (1)

wanszeto

2016-10-18 2:42 am

✔ 最佳答案

Consider the SO₂ gas produced :
P = (753.4/760) atm
V = 1.462 L
n = ? mol
R = 0.08206 L atm / (mol K)
T = (273.2 + 47.4) K = 320.6 K

PV = nRT
n = PV/RT
No. of moles of SO₂, n = (753.4/760) × 1.462 / (0.08206 × 320.6) mol = 0.05509 mol

S(s) + O₂(g) → SO₂(g)
OR: Mole ratio of S : SO₂ = 1 : 1

No. of moles of SO₂ formed = 0.05509 mol
No. of moles of S in the sample = 0.05509 mol

Molar mass of S = 32.06 g/mol
Mass of S in the sample = (32.06 g/mol) × (0.05509 mol) = 1.766 g

% by mass of S in the ore = (1.766/6.599) × 100% = 26.76%

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