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\(n_{H_2}=\frac{8,512}{22,4}=0,38\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
a <------------------------- a (mol)
2Al + 6HCl ---> 2AlCl3 + 3H2
\(\frac{2}{3}b\) <------------------------------ b (mol)
=> \(\left\{{}\begin{matrix}65a+18b=16,24\\a+b=0,38\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,18\left(mol\right)\end{matrix}\right.\)
=> mZn = 0,2.65=13(g)
=> mAl = 0,18 . \(\frac{2}{3}\) . 27 = 3,24(g)
pt Zn + 2 HCl ---> ZnCl2 +H2 (1)
2Al + 6HCl ---> 2 AlCl3 +,3H2 (2)
nH2 = 8,512/22,4 = 0,38(mol)
gọi x,y lần lượt là số mol của Zn và Al
ta có x +3y = 0,38(3)
65x + 27y = 16,24(4)
từ (3),(4) x=0,23(mol) y= 0,05(mol)
mZn = 0,23.65 = 14,95(g)
mAl = 0,05.27= 1,35(g)
Zn+2HCl--->ZnCl2+H2
x-------------------------x
2Al+6HCl--->2AlCl3+3H2
y------------------------------1,5y
n H2=8,512/22,4=0,38(mol)
Theo bài ra ta có hpt
\(\left\{{}\begin{matrix}65x+27y=16,24\\1,5x+y=0,38\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,12\end{matrix}\right.\)
m Zn=0,2.65=13(g)
m Al=16,24-13=3,24(g)
Câu 5:
PTHH : H2+ Cl2 -to-> 2 HCl
Vì số mol , tỉ lệ thuận theo thể tích , nên ta có:
25/1 = 25/1 => P.ứ hết, không có chất dư, tính theo chất nào cũng được
=> V(HCl)= 2. V(H2)= 2. 25= 50(l)
Câu 4: mFe2O3= 0,6. 80= 48(g)
=> nFe2O3= 48/160=0,3(mol)
mCuO= 80-48=32(g) => nCuO=32/80=0,4(mol)
PTHH: CuO + CO -to-> Cu + CO2
0,4_______0,4_____0,4____0,4(mol)
Fe2O3 + 3 CO -to-> 2 Fe +3 CO2
0,3_____0,9____0,6______0,9(mol)
=>nCO= 0,4+ 0,9= 1,3(mol)
=> V(CO, đktc)= 1,3. 22,4=29,12(l)
Hh: `Zn:x(mol);Al:y(mol)`
`->65x+27y=16,24(1)`
`Zn+2HCl->ZnCl_2+H_2`
`2Al+6HCl->2AlCl_3+3H_2`
Theo PT: `n_{H_2}=x+1,5y={9,4202}/{24,79}=0,38(2)`
`(1)(2)->x=0,2;y=0,12`
`m_{Zn}=0,2.65=13(g)`
`m_{Al}=16,24-13=3,24(g)`
a.\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,3 0,3 ( mol )
\(m_{Fe}=0,3.56=16,8g\)
\(\%m_{Fe}=\dfrac{16,8}{20}.100=84\%\)
\(\%m_{Cu}=100\%-84\%=16\%\)
b.\(m_{Cu}=20-16,8=3,2g\)
\(n_{Cu}=\dfrac{3,2}{64}=0,05mol\)
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,05 0,05 ( mol )
\(m_{CuO}=0,05.80=4g\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\\ Theo.pt:n_{Fe}=n_{H_2}=0,3\left(mol\right)\\ m_{Fe}=0,3.56=16,8\left(g\right)\\ m_{Cu}=20-16,8=3,2\left(g\right)\\ n_{Cu}=\dfrac{3,2}{64}=0,06\left(mol\right)\\ PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ Mol:0,05\leftarrow0,05\leftarrow0,05\\ m_{CuO}=0,05.80=4\left(g\right)\)
\(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}24a+27b=6,45\\22,4a+3.22,4b=7,28\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,15\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1.24=2,4\left(g\right)\\m_{Al}=0,15.27=4,05\left(g\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{0,953m}{22,4}=0,042545m\left(mol\right)\\ Đặt:n_{Mg}=x\left(mol\right);n_{Al}=y\left(mol\right);n_{Cu}=z\left(mol\right)\left(x,y,z>0\right)\\\Rightarrow \left\{{}\begin{matrix}24x+27y+64z=m\\40x+51y+80z=1,72m\\x+1,5y=0,042545m\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\approx0,012845m\\y\approx0,0198m\\z\approx0,002455m\end{matrix}\right.\\ \Rightarrow\%m_{Cu}\approx\dfrac{0,002455.64m}{m}.100\%\approx15,712\%\\ \%m_{Al}\approx\dfrac{27.0,0198m}{m}.100\%\approx53,46\%\\ \%m_{Mg}\approx\dfrac{0,012845.24m}{m}.100\%\approx30,828\%\)
\(n_{H_2}=\frac{8,512}{22,4}=0,38\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
a -------------------------> a (mol)
2Al + 6HCl ---> 2AlCl3 + 3H2
b ----------------------------> 1,5b (mol)
=> \(\left\{{}\begin{matrix}65a+27b=16,24\\a+1,5b=0,38\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,12\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}m_{Zn}=65.0,2=13\left(g\right)\\m_{Al}=27.0,12=3,24\left(g\right)\end{matrix}\right.\)
Mk làm câu kia r nên mk xóa câu này nha