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(A) Let the amount of mixture released each time be $y 	ext{ litres}$.The initial volume of oxygen is $16 \%$ of $8 	ext{ litres}$,which is $0.16

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$2$

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(A) Let the amount of mixture released each time be $y 	ext{ litres}$.The initial volume of oxygen is $16 \%$ of $8 	ext{ litres}$,which is $0.16 	imes 8$.After the first replacement,the remaining oxygen is $8 	imes 0.16 	imes (1 - y/8)$.After the second replacement,the remaining oxygen is $8 	imes 0.16 	imes (1 - y/8)^2$.According to the problem,the final oxygen content is $9 \%$ of the total volume $(8 	ext{ litres})$,so:$8 	imes 0.16 	imes (1 - y/8)^2 = 0.09 	imes 8$$(1 - y/8)^2 = 0.09 / 0.16$$(1 - y/8)^2 = (0.3 / 0.4)^2 = (3/4)^2$$1 - y/8 = 3/4$$y/8 = 1 - 3/4 = 1/4$$y = 8 / 4 = 2 	ext{ litres}$.

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