a,

$n_{Cu(NO_3)_2}=\dfrac{200.14,1\%}{188}=0,15(mol)$

$n_{AgNO_3}=\dfrac{200.12,75\%}{170}=0,15(mol)$

$R+nAgNO_3\to R(NO_3)_n+n Ag$

$2R+nCu(NO_3)_2\to 2R(NO_3)_n+ nCu$

Sau khi $Ag$ bị đẩy hết, $n_{Ag}=0,15(mol)$

$\Rightarrow m_{Ag}=16,2g$

Sau khi $Cu$ bị đẩy hết, $n_{Cu}=0,15(mol)$

$\Rightarrow m_{Ag}+m_{Cu}=25,8g$

Ta thấy $26,8>25,8$ nên $R$ dư, hai muối hết.

$m_{R\text{dư}}=26,8-25,8=1g$

$\Rightarrow m_{R\text{pứ}}=5,05-1=4,05g$

$n_R=\dfrac{0,15}{n}+\dfrac{0,15.2}{n}=\dfrac{0,45}{n}(mol)$

$\Rightarrow M_R=\dfrac{4,05n}{0,45}=9n$

$n=3\Rightarrow M_R=27(Al)$

b,

$m_{dd\text{spu}}=4,05+200-25,8=178,25g$

$n_{Al(NO_3)_3}=n_{Al}=0,15(mol)$

$\Rightarrow C\%_{Al(NO_3)_3}=\dfrac{0,15.213.100}{178,25}=17,92\%$

Đáp án:a)Al b)17.92%

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