\(n_{Mg}=\dfrac{3,6}{24}=0,15mol\\ Mg+2HCl\rightarrow MgCl_2+H_2\)
0,15 0,3 0,15 0,15
\(V_{H_2}=0,15.22,4=3,36l\\ b)m_{ddHCl}=\dfrac{0,3.36,5}{5}\cdot100=219g\\ b)C_{\%MgCl_2}=\dfrac{0,15.95}{3,6+210-0,15.2}\cdot100=6,68\%\)
\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}=\dfrac{3,6}{24}\approx0,15mol\)
PTHH: Mg + 2HCl \(\rightarrow\) MgCl2 + H2
TL: 1 : 2 : 1 : 1
mol: 0,15 \(\rightarrow\) 0,3 \(\rightarrow\) 0,15 \(\rightarrow\) 0,15
\(a.V_{H_2}=n_{H_2}.22,4=0,15.22,4=3,36l\)
\(b.m_{HCl}=n_{HCl}.M_{HCl}=0,15.36,5=5,5g\)
\(C\%_{ddHCl}=\dfrac{m_{HCl}}{m_{ddHCL}}.100\%\)
\(\Leftrightarrow5\%=\dfrac{5,5}{m_{ddHCl}}.100\%\)
\(\Leftrightarrow m_{ddHCl}=\dfrac{100\%.5,5}{5\%}\)
\(\Rightarrow m_{ddHCl}=110g\)
\(m_{MgCl_2}=n.M=0,15.95=14,3g\)
\(c.m_{ddspu}=m_{Mg}+m_{ddHCl}=3,6+110=113,6g\)
\(C\%_{ddspu}=\dfrac{m_{MgCl_2}}{m_{ddspu}}.100\%=\dfrac{14,3}{113,6}.100\%\approx12,6\%\)
