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Stoichiometric Calculations

The weight of...

Question

The weight of $S O_{2}$ formed when 20 g of sulphur is burnt in excess of $O_{2}$ is:

32 g

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64 g

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40 g

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60 g

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Solution

The correct option is C 40 g
We have,

$S + O_{2} \to S O_{2}$

moles of sulphur = $\frac{20}{32} = 0.625$

0.625 mole of Sulphur will form 0.625moles of $S O_{2}$

so, weight of $S O_{2}$ = $0.625 \times 64 = 40 g$

Option C is correct.

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Similar questions

S + O_{2} \to S O_{2} (Yield = 80%)

S O_{2} + C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l (Yield = 60%)

H_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 H C l (Yield = 30%)

S O_{2}

C l_{2}

B a C l_{2}

B a S O_{4}

137 g/mol

4

(H_{2})

64

44.8

O_{2}

H_{2} S O_{4}

49

H_{2} S O_{4}

%

{S O}_{2}

S = 32

O = 16

20

:

1.0 g

0.56 g O_{2}

(At. wt. Mg = 24; O = 16)

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