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(B) Step $1$: Calculate the molar masses of the substances: $M(SO_2) = 32 + 2 	imes 16 = 64 \ g/mol$ $M(O_2) = 2 	imes 16 = 32 \ g/mol$ $M(SO_3) = 3

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$30$

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(B) Step $1$: Calculate the molar masses of the substances: $M(SO_2) = 32 + 2 	imes 16 = 64 \ g/mol$ $M(O_2) = 2 	imes 16 = 32 \ g/mol$ $M(SO_3) = 32 + 3 	imes 16 = 80 \ g/mol$ Step $2$: Calculate the number of moles at equilibrium: $n(SO_2) = \frac{12.8 \ g}{64 \ g/mol} = 0.2 \ mol$ $n(O_2) = \frac{9.6 \ g}{32 \ g/mol} = 0.3 \ mol$ $n(SO_3) = \frac{48 \ g}{80 \ g/mol} = 0.6 \ mol$ Step $3$: Since the volume is $1 \ L$,the concentrations are equal to the number of moles: $[SO_2] = 0.2 \ M, [O_2] = 0.3 \ M, [SO_3] = 0.6 \ M$ Step $4$: Calculate $K_c$ using the equilibrium expression: $K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]} = \frac{(0.6)^2}{(0.2)^2 	imes 0.3} = \frac{0.36}{0.04 	imes 0.3} = \frac{0.36}{0.012} = 30$

If the volume of the container is $1 \ L$ and at equilibrium the amounts are $SO_3 = 48 \ g$,$SO_2 = 12.8 \ g$,and $O_2 = 9.6 \ g$,find the value of $K_c$ for the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.

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