Galena (an ore) is partially oxidized by pass | vedclass

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(C) The chemical reaction for the self-reduction process of Galena $(PbS)$ is: $PbS + O_2 \rightarrow Pb + SO_2$ From the stoichiometry of the balance

Vedclass · Accepted Answer

$6.47$

Vedclass · Answer

(C) The chemical reaction for the self-reduction process of Galena $(PbS)$ is: $PbS + O_2 ightarrow Pb + SO_2$ From the stoichiometry of the balanced equation,$1 \ mol$ of $O_2$ produces $1 \ mol$ of $Pb$. Therefore,the number of moles of $O_2$ consumed equals the number of moles of $Pb$ produced: $n_{O_2} = n_{Pb}$ Using the relation $n = \frac{w}{M}$,we get: $\frac{w_{O_2}}{M_{O_2}} = \frac{w_{Pb}}{M_{Pb}}$ Given $M_{O_2} = 2 	imes 16 = 32 \ g \ mol^{-1}$ and $M_{Pb} = 207 \ g \ mol^{-1}$. For $w_{O_2} = 1 \ kg$: $w_{Pb} = \frac{w_{O_2} 	imes M_{Pb}}{M_{O_2}} = \frac{1 \ kg 	imes 207}{32} = 6.46875 \ kg \approx 6.47 \ kg$.

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