Đáp án + Giải thích các bước giải:

Câu 13:

$PTHH:$

$2Na+2H_2O\to 2NaOH+H_2\uparrow(1)$

$2Al+2NaOH+2H_2O\to 2NaAlO_2+3H_2\uparrow(2)$

$n_{H_2(1)}=\dfrac{4,48}{22,4}=0,2(mol)$

$n_{H_2(2)}=\dfrac{5,152}{22,4}=0,23(mol)$

Sau phản ứng với nước thu được chất rắn không tan gồm $Al,Fe$

$\Rightarrow m_{Al}+m_{Fe}=6(g)$

Theo PT(1): $n_{Na}=2n_{H_2(1)}=0,4(mol)$

$\Rightarrow m_{Na}=0,4.23=9,2(g)$

Theo PT(2): $n_{Al}=\dfrac{2}{3}n_{H_2(2)}=\dfrac{23}{150}(mol)$

$\Rightarrow m_{Al}=\dfrac{23}{150}.27=4,14(g)$

$\Rightarrow m_{Fe}=6-4,14=1,86(g)$

$\Rightarrow m_{\text{hỗn hợp}}=1,86+4,14+9,2=15,2(g)$

$\Rightarrow \begin{cases} \%m_{Al}=\dfrac{4,14}{15,2}.100\%=27,24\%\\ \%m_{Fe}=\dfrac{1,86}{15,2}.100\%=12,24\%\\ \%m_{Na}=100-12,14-27,24=60,62\% \end{cases}$

Câu 14:

$PTHH:$

$2Na+2H_2O\to 2NaOH+H_2\uparrow(1)$

$2Al+2NaOH+2H_2O\to 2NaAlO_2+3H_2\uparrow(2)$

$n_{H_2(1)}=\dfrac{5,6}{22,4}=0,25(mol)$

$n_{H_2(2)}=\dfrac{12,32}{22,4}=0,55(mol)$

Theo PT(1): $n_{Na}=2n_{H_2(1)}=0,5(mol)$

$\Rightarrow m_{Na}=0,5.23=11,5(g)$

Theo PT(2): $n_{Al}=\dfrac{2}{3}n_{H_2(2)}=\dfrac{11}{30}(mol)$

$\Rightarrow m_{Al}=\dfrac{11}{30}.27=9,9(g)$

$\Rightarrow m_X=m_{Al}+m_{Na}=9,9+11,5=21,4(g)$

$\to m=21,4$