nH2=\(\dfrac{3,92}{22,4}\)=0,175(mol)
ZnO+H2--to-->Zn+H2O
x_____x_____x___x
CuO+H2-to-->Cu+H2O
y____y______y___y
ta có hệ \(\left\{{}\begin{matrix}81x+80y=14,1\\x+y=0,175\end{matrix}\right.\)
=>x=0,1 mol, y=0,075 mol
=>m chất rắn=x=0,1.65+0,075.64=11,3(g)
=>m H2O=y=18(0,1+0,075)=3,15(g)
a: \(ZnO+H_2\rightarrow Zn+H_2O\)
x x
\(CuO+H_2\rightarrow Cu+H_2O\)
y y
b: \(n_{H_2}=\dfrac{3.92}{22.4}=0.175\left(mol\right)\)
Theo đề, ta có: x+y=0,175 và 81x+80y=14,1
=>x=0,1; y=0,075
=>\(n_{Zn}=0.1mol;n_{Cu}=0.075mol\)
\(m=0.1\cdot65+0.075\cdot64=11.3\left(g\right)\)
