Mg + 2HCl → MgCl2 + H2 (1)
CuO + 2HCl → CuCl2 + H2O (2)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,4\times2=0,8\left(g\right)\)
a) Theo PT1: \(n_{HCl}=2n_{H_2}=2\times0,4=0,8\left(mol\right)\)
Theo PT1: \(n_{Mg}=n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,4\times24=9,6\left(g\right)\)
\(\Rightarrow m_{CuO}=17,6-9,6=8\left(g\right)\)
\(\Rightarrow n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
Theo PT2: \(n_{HCl}=2n_{CuO}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,8+0,2=1\left(mol\right)\)
\(\Rightarrow m_{HCl}=1\times36,5=36,5\left(g\right)\)
\(\Rightarrow C\%_{HCl}=\dfrac{36,5}{300}\times100\%=12,17\%\)
b) Theo PT1: \(n_{MgCl_2}=n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=0,4\times95=38\left(g\right)\)
Theo PT2: \(n_{CuCl_2}=n_{CuO}=0,1\left(mol\right)\)
\(\Rightarrow m_{CuCl_2}=0,1\times135=13,5\left(g\right)\)
\(\Rightarrow m_{muối}=13,5+38=51,5\left(g\right)\)
\(m_{dd}saupư=17,6+300-0,8=316,8\left(g\right)\)
\(\Rightarrow C\%_{muối}saupư=\dfrac{51,5}{316,8}\times100\%=16,26\%\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\\ 0,4mol:0,8mol\leftarrow0,4mol:0,4mol\)
\(CuO+2HCl\rightarrow H_2O+CuCl_2\\ 0,1mol:0,2mol\rightarrow0,1mol:0,1mol\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
\(m_{Mg}=0,4.24=9,6\left(g\right)\)
\(\Leftrightarrow m_{CuO}=17,6-9,6=8\left(g\right)\)
\(\Leftrightarrow n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
\(n_{HCl}=0,8+0,2=1\left(mol\right)\)
\(m_{HCl}=36,5.1=36,5\left(g\right)\)
a. \(C\%HCl=\dfrac{36,5}{300}.100\%=12\%\)
b. \(m_{muoi}=m_{MgCl_2}+m_{CuCl_2}\)
\(\Leftrightarrow m_{muoi}=0,4.95+0,1.135\)
\(\Leftrightarrow m_{muoi}=38+13,5=51,5\left(g\right)\)
\(m_{H_2}=2.0,4=0,8\left(g\right)\)
\(m_{H_2O}=18.0,1=1,8\left(g\right)\)
\(m_{ddsaupu}=17,6+300-\left(0,8+1,8\right)=315\left(g\right)\)
\(\Leftrightarrow C\%muoisaupu=\dfrac{51,5}{315}.100\%=16,35\%\)
\(\left(?\right)\)
