\(n_{Na}=\dfrac{6,9}{23}=0,3mol\)
\(n_{HCl}=0,1.0,1=0,01mol\)
\(2Na+2HCl\rightarrow2NaCl+H_2\)
0,3 < 0,01 ( mol )
0,01 0,01 0,01 ( mol )
\(m_{HCl}=0,01.36,5=0,365g\)
\(m_{NaCl}=0,01.58,5=0,585g\)
\(m_{ddspứ}=0,365+6,9-0,01.2=7,245g\)
\(C\%_{NaCl}=\dfrac{0,585}{7,245}.100=8,07\%\)
\(C_{M_{NaCl}}=\dfrac{0,01}{0,1}=0,1M\)