Quay lại đáp án

$n_{OH-}$$=$$0,3$$.$$0,2$$+$$0,025$$.$$0,2$$.$$2$$=$$0,07$$n_{Al3+}$$=$$0,1$$.$$2$$.$$0,2$$=$$0,04$ $Al^{3+}$$+$$3$$OH^{-}$$\rightarrow $$Al(OH)_{3}$ $\frac{0,07}{3}$$\leftarrow$$0,07$$\rightarrow$$\frac{0,07}{3}$$n_{Ba^{2+}}$$=$0,025.0,2=0,005$n_{SO4^{2-}}$$=$3.0,1.0,2=0,06 $Ba^{2+}$$+$$SO4^{2-}$$\rightarrow$$BaSO4$ $0,003$$\rightarrow$ $0,003$$m$$=$ $\frac{0,07}{3}$$.$$78$$+$$0,003$$.$$233$$=$$2.519$bài có gì sai bạn bảo lại m nha

$n_{OH-}$$=$$0,3$$.$$0,2$$+$$0,025$$.$$0,2$$.$$2$$=$$0,07$$n_{Al3+}$$=$$0,1$$.$$2$$.$$0,2$$=$$0,04$ $Al^{3+}$$+$$3$$OH^{-}$$\rightarrow $$Al(OH)_{3}$ $\frac{0,07}{3}$$\leftarrow$$0,07$$\rightarrow$$\frac{0,07}{3}$$n_{Ba^{2+}}$$=$0,025.0,2=0,005$n_{SO4^{2-}}$$=$3.0,1.0,2=0,06 $Ba^{2+}$$+$$SO4^{2-}$$\rightarrow$$BaSO4$ $0,005$$\rightarrow$ $0,005$$m$$=$ $\frac{0,07}{3}$$.$$78$$+$$0,005$$.$$233$$=$$2.985$bài có gì sai bạn bảo lại m nha