\(2Fe+3Cl2-->2FeCl3\)(1)
\(Cu+Cl2--->CuCl2\)(2)
\(Fe+2HCl-->FeCl2+H2\)(3)
\(n_{FeCl2}=\frac{25,4}{127}=0,2\left(mol\right)\)
Theo pthh3
\(n_{Fe}=n_{FeCl2}=0,2\left(mol\right)\)
\(m_{Fe}=0,2.56=11,2\left(g\right)\)
Theo pthh1
\(n_{FeCl2}=n_{Fe}=0,2\left(mol\right)\)
\(m_{FeCl2}=0,2.127=25,4\left(g\right)\)
\(m_{CuCl2}=59,5-25,4=34,1\left(g\right)\)
\(n_{CuCl2}=\frac{34,1}{135}=0,253\left(mol\right)\)
\(n_{Cu}=n_{CuCl2}=0,253\left(mol\right)\)
\(m_{Cu}=0,253.64=16,192\left(g\right)\)
\(n_{HCl}=n_{Fe}=0,2\left(mol\right)\)
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(m_{ddHCl}=\frac{7,3.100}{10}=73\left(g\right)\)
\(V_{HCl}=\frac{73}{1,049}=69,59\left(l\right)\)
2Fe+3Cl2−−>2FeCl3(1)
Cu+Cl2−−−>CuCl2(2)
Fe+2HCl−−>FeCl2+H2(3)
nFeCl2=25,4127=0,2(mol)
Theo pthh3
nFe=nFeCl2=0,2(mol)
mFe=0,2.56=11,2(g)
Theo pthh1
nFeCl2=nFe=0,2(mol)
mFeCl2=0,2.127=25,4(g)
mCuCl2=59,5−25,4=34,1(g)
nCuCl2=34,1135=0,253(mol)
nCu=nCuCl2=0,253(mol)
mCu=0,253.64=16,192(g)
nHCl=nFe=0,2(mol)
mHCl=0,2.36,5=7,3(g)
mddHCl=7,3.10010=73(g)
VHCl=731,049=69,59(l)
