$n_{NO_2}=\dfrac{5,376}{22,4}=0,24(mol)$

$Al+6HNO_3\to Al(NO_3)_3+3NO_2\uparrow+3H_2O$

$Al_2O_3+6HNO_3\to 2Al(NO_3)_3+3H_2O$

$\Rightarrow n_{Al}=\dfrac{1}{3}n_{NO_2}=0,08(mol)$

$\Rightarrow m_{Al}=0,08.27=2,16(g)$

$\Rightarrow \%_{Al_2O_3}=\dfrac{14,5-2,16}{14,5}.100\%\approx 85,1\%$

$n_{Al_2O_3}=\dfrac{14,5-2,16}{102}\approx 0,12(mol)$

`\Rightarrow \sum n_{Al(NO_3)_3}=n_{Al}+2n_{Al_2O_3}=0,32(mol)`

$\Rightarrow C_{M_{Al(NO_3)_3}}=\dfrac{0,32}{0,8}=0,4M$

Đáp án:

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Giải thích các bước giải:

$n_{NO_2} = {5,376}/{22,4} = 0,24(mol) \\ PTHH : \\ Al + 6HNO_3 \to Al(NO_3)_3 + 3NO_2 + 3H_2O \\  Al_2O_3 + 6HNO_3 \to 2Al(NO_3)_3 + 3H_2O \\ n_{Al} = 1/3n_{NO_2} = 0,08(mol) \\ m_{Al} = 0,08.27 = 2,16 \\ m_{Al_2O_3} = 14,5 - 2,16 = 12,34(gam) \\ %m_{Al_2O_3} = {12,34}/{14,5}.100% = 85,1% \\ n_{Al_2O_3} = {12,34}/{102} = 0,12(mol) \\ n_{Aln_{Al(NO_3)_3}= 0,08+0,12.2 = 0,32(mol) \\ CM_{Al(NO_3)_3} = \dfrac{0,32}{0,8} = 0,3(M)$