Gọi số mol: \(\left\{{}\begin{matrix}n_{Cu}:x\left(mol\right)\\n_{Mg}:y\left(mol\right)\\n_{Fe}:z\left(mol\right)\end{matrix}\right.\)
\(64x+24y+56z=16,8\left(1\right)\)
\(135x+95y+162.5z=48,75\left(2\right)\)
Thí nghiệm 2:
Gọi k là hệ số liên hệ giũa 2 thí nghiệm
Ta có :
\(kx+ky+kz=0,8\left(3\right)\)
Đổi 600ml = 0,6l
\(n_{HCl}=0,6.2=1,2\left(mol\right)\)
\(2n_{H2}=n_{HCl}\Rightarrow n_{H2}=\frac{1,2}{2}=0,6\left(mol\right)\)
\(2ky+2kz=2n_{H2}=0,6.2=1,2\left(4\right)\)
(3)/(4) để triệt k
Ta được: \(\frac{k\left(x+y+z\right)}{k\left(2y+2z\right)}=\frac{0,8}{1,2}\)
\(\Leftrightarrow\frac{x+y+z}{2y+2z}=\frac{2}{3}\)
\(\Leftrightarrow3x+3y=3z=4y+4z\)
\(\Leftrightarrow3x-y-z=0\left(5\right)\)
\(\left(1\right)+\left(2\right)+\left(5\right)\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\\z=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Cu}=0,1\left(mol\right)\\n_{Mg}=0,2\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)
