How many $O_{2}$ $O_2$ 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 $S (I V)$ $S(IV)$ oxide, $S O_{2} (g)$ $SO_2(g)$

$S + O_{2} (g) \to S O_{2} (g)$ $S + O_2(g) rarr SO_2(g)$

Moles of $S$ $S$ $=$ $=$ $\frac{7.87 \cdot g}{32.06 \cdot g \cdot m o l^{- 1}}$ $(7.87*g)/(32.06*g*mol^-1)$ $=$ $=$ $0.245 \cdot m o l$ $0.245*mol$

And thus, by reaction stoichiometry, we need $0.245 \cdot m o l$ $0.245*mol$ dioxygen gas.

So we need, $0.245 \cdot m o l \times 6.022 \times 10^{23} \cdot m o l^{- 1} oxygen molecules$ $0.245*molxx6.022xx10^23*mol^-1" oxygen molecules"$ .

How many oxygen atoms does this represent?

How many O2O_2 molecules are needed to react with 7.87 g of S?