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Question

chia hỗn hợp X gồm Na, Al, Mg làm 3 phần bằng nhau

phần 1 tác dụng vs nước sinh ra 8.96 l khí

phần 2 tác dụng vs NaOH dư thì thấy sinh ra 15.68 lít khí

phần 3 tác dụng vs dung dịch HCL phản ứng xon...

Trần Quốc Lộc · Accepted Answer

a) $P_1:2Na+2H_2O\rightarrow2NaOH+H_2\left(1\right)$

$P_2:2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\left(2\right)$

$P_3:2Na+2HCl\rightarrow2NaCl+H_2\left(3\right)\
2Al+6HCl\rightarrow2AlCl_3+3H_2\left(4\right)\
Mg+2HCl\rightarrow MgCl_2+H_2\left(5\right)$

$b\text{) }n_{H_2\left(1\right)}=\dfrac{V}{22,4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\ n_{H_2\left(2\right)}=\dfrac{V}{22,4}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\ n_{H_2\left(3;4;5\right)}=\dfrac{V}{22,4}=\dfrac{26,88}{22,4}=1,2\left(mol\right)$

Theo $pthh\left(1\right):n_{Na}=2n_{H_2\left(1\right)}=2\cdot0,4=0,8\left(mol\right)$

Theo $pthh\left(2\right):n_{Al}=\dfrac{2}{3}n_{H_2\left(2\right)}=\dfrac{2}{3}\cdot0,7=\dfrac{7}{15}\left(mol\right)$

Theo $pthh\left(3\right):n_{H_2\left(3\right)}=n_{Na}=\dfrac{1}{2}\cdot0,8=0,4\left(mol\right)$

Theo $pthh\left(4\right):n_{H_2\left(4\right)}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\cdot\dfrac{7}{15}=0,7\left(mol\right)$

$\Rightarrow n_{H_2\left(5\right)}=n_{H_2\left(3;4;5\right)}-n_{H_2\left(3;4\right)}=1,2-\left(0,7+0,4\right)=0,1\left(mol\right)$

Theo $pthh\left(5\right):n_{Mg}=2n_{H_2\left(5\right)}=2\cdot0,1=0,2\left(mol\right)$

$\Rightarrow m_{Al}=n\cdot M=3\cdot\dfrac{7}{15}\cdot27=37,8\left(g\right)\ m_{Na}=n\cdot M=3\cdot0,8\cdot23=55,2\left(g\right)\
m_{Mg}=n\cdot M=3\cdot0,2\cdot24=14,4\left(g\right)$Câu trả lời: a) $P_1:2Na+2H_2O\rightarrow2NaOH+H_2\left(1\right)$