What mass of sulfur has to burn to produce 4.5L SO2 at 300°C and 101 kPa in the following reaction?s(s)+O2(g) SO,(g)
1 answer:
<span>Answer: option B. 3.07 g
Explanation:
1) given reaction:
S(s) + O₂ (g) → SO(g)
2) Balanced chemical equation:
</span><span>2S(s) + O₂ (g) → 2SO(g)
3) Theoretical mole ratios:
2 mol S : 1 mol O₂ : 2 mol SO
3) number of moles of 4.5 liter SO₂ at</span><span> 300°C and 101 kPa
use the ideal gas equation:
pV = nRT
with V = 4.5 liter
p = 101 kPa
T = 300 + 273.15 K = 573.15 K
R = 8.314 liter×kPa / (mol×K)
=> n = pV / (RT) =
n = [101 kPa × 4.5 liter] / [8.314 (liter×kPa) / (mol×K) × 573.15 K ]
n = 0.0954 mol SO
4) proportion with the theoretical ratio S / SO
2 mol S x
-------------- = ----------------------
2 mol SO 0.0954 mol SO
=> x = 0.0954 mol S.
5) Convert mol of S to grams by using atomic mass of S = 32.065 g/mol
mass = number of moles × atomic mass
mass = 0.0954 mol × 32.065 g/mol = 3.059 g of S
6) Therefore the answer is the option B. 3.07 g
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