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Stoichiometric Calculations

Sulphur triox...

Question

Sulphur trioxide may be prepared by the following two reactions:

$S_{8} + {8 O}_{2} (g) ⟶ {8 S O}_{2} (g)$
${2 S O}_{2} (g) + O_{2} (g) ⟶ {2 S O}_{3} (g)$

How many grams of sulphur trioxide will be produced from 1 mol of $S_{8}$ ?

60g

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640g

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80g

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512g

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Solution

The correct option is B 640g
Given that
$S_{8} + 8 O_{2} \to 8 S O_{2}$ ____(I)
$2 S O_{2} + O_{2} \to 2 S O_{3}$ ____(II)
Now, $(I) + 4 \times (I I)$
$S_{8} + 8 O_{2} + 8 S O_{2} + 4 O_{2} \to 8 S O_{2} + 8 S O_{3}$

$S_{8} + 12 O_{2} \to 8 S O_{3}$
$1$ mol $8$ mol
$8 \times 32 g$ _____ $8 \times 80 g$
$∴$ mass of $S O_{3}$ formed $= 8 \times 80$
$= 640 g$

$(B) 640 g$

Suggest Corrections

Similar questions

S_{8} + 8 O_{2} (g) \to 8 S O_{2} (g)

2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g)

S O_{3}

1

S_{8}

S_{8} (s) + 8 O_{2} (g) \to 8 S O_{2} (g)

2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g)

S O_{3}

S_{8}

S_{8} (s) + 8 O_{2} (g) \to 8 S O_{2} (g)

2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g)

S O_{3}

1

S_{8}

S_{8} (s) + 8 O_{2} (g) \to 8 S O_{2} (g); Δ H = - 2374. k J

2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g); Δ H = - 198. k J

S_{8} (s) + 80_{2} (g) \to {8 S O}_{2} (g)

{2 S O}_{8} (g) + O_{2} (g) \to {2 S O}_{3} (g)

{S O}_{3}

1

{S O}_{8}

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