Question #0cbe1

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Question #0cbe1

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Answer 1 · 2018-01-14T12:23:33

The tequired mass of sulfur is $25 g$ $25"g"$ .

Explanation:

We have a very simple chemical reaction equation:

$S + O_{2} \to {SO}_{2}$ $"S"+"O"_2\rightarrow "SO"_2$

This the ratio of reactant to product is one mole of sulfur to one mole of sulfur dioxide. But next we have to find the total number of moles corresponding to $50 g$ $50"g"$ of sulfur dioxide.

To do that add p the atomic weights to get the molecular weight:

$(\frac{32 g S}{mol}) + 2 \times (\frac{16 g O}{mol}) = (\frac{64 {g SO}_{2}}{mol})$ $((32"g S")/"mol")+2×((16"g O")/"mol")=((64"g SO"_2)/"mol")$

And then divide this into the given mass of sulfur dioxide:

$\frac{50 {g SO}_{2}}{(\frac{64 {g SO}_{2}}{mol})} = 0.78 {mol SO}_{2}$ $(50"g SO"_2)/(((64"g SO"_2)/"mol"))=0.78"mol SO"_2$

Only two significant digits are used because, in fact, the given atomic weight of sulfur is correct only to that accuracy (four significant digits would be $32.06$ $32.06$ ).

So we have $0.78$ $0.78$ mole of sulfur dioxide meaning there must be $0.78$ $0.78$ of sulfur. Then multiplying by the atomic weight gives the required mass of sulfur:

$(0.78 mol S) \times \frac{32 g}{mol} = 25 g S$ $(0.78"mol S")×(32"g")/("mol")=25"g S"$

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