20 g of iron pyrite,FeS_2,is roasted complet | vedclass

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(B) Step $1$: Calculate moles of $FeS_2$. Molar mass of $FeS_2 = 56 + 2 	imes 32 = 120 \ g/mol$. Moles of $FeS_2 = \frac{20 \ g}{120 \ g/mol} = \frac

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$\frac{10}{3} \ M$

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(B) Step $1$: Calculate moles of $FeS_2$. Molar mass of $FeS_2 = 56 + 2 	imes 32 = 120 \ g/mol$. Moles of $FeS_2 = \frac{20 \ g}{120 \ g/mol} = \frac{1}{6} \ mol$.Step $2$: Write the roasting reaction: $2FeS_2 + \frac{11}{2}O_2 	o Fe_2O_3 + 4SO_2$.From stoichiometry,$2 \ mol$ of $FeS_2$ produce $4 \ mol$ of $SO_2$. Thus,$\frac{1}{6} \ mol$ of $FeS_2$ produces $\frac{1}{6} 	imes 2 = \frac{1}{3} \ mol$ of $SO_2$.Step $3$: Write the absorption reaction: $SO_2 + 2NaOH 	o Na_2SO_3 + H_2O$.To react with $\frac{1}{3} \ mol$ of $SO_2$,we need $\frac{2}{3} \ mol$ of $NaOH$.Step $4$: Calculate molarity. Given that $50\%$ of the $NaOH$ solution was used,let $M$ be the molarity. Total moles of $NaOH = \frac{400 	imes M}{1000} = 0.4M$.$0.5 	imes 0.4M = \frac{2}{3} \ mol$ $\Rightarrow 0.2M = \frac{2}{3}$ $\Rightarrow M = \frac{2}{0.6} = \frac{10}{3} \ M$.

$20 \ g$ of iron pyrite,$FeS_2$,is roasted completely and the $SO_2$ gas produced is absorbed completely in $400 \ mL$ of $NaOH$ solution. If only $50\%$ of the $NaOH$ is used in the reaction,what was the molarity of the $NaOH$ solution?

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