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Sulphur

Sulphur triox...

Question

Sulphur trioxide may be prepared by the following two reactions:
$S_{8} + 8 O_{2} (g) \to 8 S O_{2} (g)$
$2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g)$ .
How many grams of $S O_{3}$ will be produced from $1$ mol of $S_{8}$ ?

640.0

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678.5

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750.3

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775.8

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Solution

The correct option is A $640.0$ g
From both the reactions, it can be seen that $1$ mole reactant will produce $8$ mole of $S O_{3} = 80 \times 8 = 640$ g.

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Similar questions

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

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

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)

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8 F e + S_{8} \to 8 F e S

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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)

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{S O}_{8}

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