How many $O_2$ molecules are needed to react with 7.87 g of S?

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Answer 1 · 2016-08-18T05:35:09

Wee assume formation of the $S(IV)$ oxide, $SO_2(g)$

$S + O_2(g) rarr SO_2(g)$

Moles of $S$ $=$ $(7.87*g)/(32.06*g*mol^-1)$ $=$ $0.245*mol$

And thus, by reaction stoichiometry, we need $0.245*mol$ dioxygen gas.

So we need, $0.245*molxx6.022xx10^23*mol^-1" oxygen molecules"$ .

How many oxygen atoms does this represent?

How many O_2O_2 molecules are needed to react with 7.87 g of S?