Ta có: mH2O = mchất rắn giảm= 5,4(g)
=> mChất rắn= 21,5- 5,4= 16,1(g)
-> nH2O= 5,4/18= 0,3(mol)
- Gọi: nMg(OH)2= x (mol); nZn(OH)2= y(mol)
PTHH: (1) Mg(OH)2 -to-> MgO + H2O
_________x___________________x (mol)
(2) Zn(OH)2 -to-> ZnO + H2O
____y_________________y (mol)
=> 58x + 99y= 21,5
x+y= 0,3
=> x= 0,2; y= 0,1
=> nMg(OH)2= 0,2(mol); nZn(OH)2= 0,1(mol)
=> mMg(OH)2= 0,2.58= 11,6(g)
=> %mMg(OH)2= (11,6/21,5).100 ~ 53,953%
=> %mZn(OH)2 ~ 100% - 53,953% ~ 46, 047%
CHÚC BẠN HỌC TỐT!!
Gọi x, y lần lượt là số mol của Mg(OH)2 và Zn(OH)2
PTHH: \(Mg\left(OH\right)_2\underrightarrow{O}MgO+H_2O\)
pư...............x.................x..............x(mol)
PTHH: \(Zn\left(OH\right)_2\underrightarrow{O}ZnO+O_2\)
pư...............y.................y...........y(mol)
Theo đề bài, ta có:
+)\(m_{Mg\left(OH\right)2}+m_{Zn\left(OH\right)2}=21,5\left(g\right)\)(1)
+) \(m_{MgO}+m_{ZnO}=m_{hhX}-5,4\)
\(\Rightarrow m_{MgO}+m_{ZnO}=21,5-5,4=16,1\left(g\right)\)(2)
Từ (1)(2) suy ra: \(\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}m_{MgO}=40.0,2=8\left(g\right)\\m_{ZnO}=16,1-8=8,1\left(g\right)\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}m_{Mg\left(OH\right)2}=58.0,2=11,6\left(g\right)\\m_{Zn\left(OH\right)2}=21,5-11,6=9,9\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%Mg\left(OH\right)_2=\dfrac{11,6}{21,5}.100\%\approx53,95\%\\\%Zn\left(OH\right)_2=\dfrac{9,9}{21,5}.100\%\approx46,05\%\end{matrix}\right.\)
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