Gọi \(\left\{{}\begin{matrix}n_R=a\left(mol\right)\\n_{RO}=b\left(mol\right)\\n_{RCO_3}=c\left(mol\right)\end{matrix}\right.\left(ĐK:a,b,c>0\right)\)
`=>` \(aM_R+b\left(M_R+16\right)+c\left(M_R+60\right)=60,45\left(1\right)\)
Ta có: \(\left\{{}\begin{matrix}n_{H_2SO_4}=\dfrac{375.19,6\%}{98}=0,75\left(mol\right)\\n_{khí}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\end{matrix}\right.\)
PTHH: \(R+H_2SO_4\rightarrow RSO_4+H_2\) (1)
a----->a----------->a------>a
\(RO+H_2SO_4\rightarrow RSO_4+H_2O\) (2)
b------>b----------->b
\(RCO_3+H_2SO_4\rightarrow RSO_4+CO_2\uparrow+H_2O\)
c--------->c------------>c-------->c
`=>` \(\left\{{}\begin{matrix}\left(a+b+c\right)\left(M_R+96\right)=104,65\left(2\right)\\a+c=0,45\\n_{H_2SO_4\left(dư\right)}=0,75-a-b-c\left(mol\right)\end{matrix}\right.\)
Ta có: \(m_{ddspư}=375+60,45-2a-44c=435,45-2a-44c\left(g\right)\)
`=>` \(C\%_{H_2SO_4\left(dư\right)}=\dfrac{98\left(0,75-a-b-c\right)}{435,45-2a-44c}.100\%=2,311\%\left(4\right)\)
Từ \(\left(1\right),\left(2\right),\left(3\right),\left(4\right)\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,2\\c=0,25\\M_R=65\left(g/mol\right)\end{matrix}\right.\)
`=> A: Zn`
`=>` \(A:\left\{{}\begin{matrix}m_{Zn}=0,2.65=13\left(g\right)\\m_{ZnO}=0,2.81=16,2\left(g\right)\\m_{ZnCO_3}=0,25.125=31,25\left(g\right)\end{matrix}\right.\)
