\(n_{HCl}=\dfrac{3,65\%.100}{100\%.36,5}=0,1\left(mol\right)\)
Pt : \(2Na+2HCl\rightarrow2NaCl+H_2\)
0,15 0,1 0,1 0,05
Xét tỉ lệ : \(\dfrac{0,15}{2}>\dfrac{0,1}{2}\Rightarrow Nadư\)
\(m_{ddspu}=0,15.23+100-0,05.2=103,35\left(g\right)\)
\(C\%_{NaCl}=\dfrac{0,1.58,5}{103,35}.100\%=5,66\%\)
Chúc bạn học tốt
\(n_{HCl}=\dfrac{100.3,65}{100}:3,65=0,1mol\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
0,15 0,15 0,075
\(NaOH+HCl\rightarrow NaCl+H_2O\\ \Rightarrow\dfrac{0,15}{1}>\dfrac{0,1}{1}\Rightarrow NaOH.dư\\ n_{HCl}=n_{NaOH}=n_{NaCl}=0,1mol\\ m_{dd}=0,15.23+100-0,075.2=103,3g\\ C_{\%NaCl}=\dfrac{0,1.58,5}{103,3}\cdot100=5,66\%\\ C_{\%NaOH\left(dư\right)}=\dfrac{\left(0,15-0,1\right).40}{103,3}\cdot100=1,94\%\)
