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a) \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
0,3--------------->0,3--->0,3
=> \(m_{MgCl_2}=0,3.95=28,5\left(g\right)\)
b)
\(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c)
\(n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,3}{1}\) => CuO dư, H2 hết
PTHH: CuO + H2 --to--> Cu + H2O
0,3<--0,3------->0,3
=> mchất rắn = 32 - 0,3.80 + 0,3.64 = 27,2 (g)
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Mol : 0,4 0,4 0,6
\(m_{AlCl_3}=133,5.0,4=53,4\left(g\right)\\ V_{H_2}=0,6.22,4=13,44\left(l\right)\)
\(n_{CuO}=\dfrac{8}{80}=0,8\left(mol\right)\\
pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\
LTL:\dfrac{0,8}{1}>\dfrac{0,6}{1}\)
=> CuO dư
\(n_{CuO\left(p\text{ư}\right)}=n_{Cu}=n_{H_2}=0,6\left(mol\right)\\
m_{CuO\left(d\right)}=\left(0,8-0,6\right).80=16\left(g\right)\\
m_{Cu}=0,6.64=38,4\left(g\right)\\
m_{cr}=16+38,4=54,4\left(g\right)\)
a) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,4-------------->0,4-->0,6
=> \(m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\)
b) \(V_{H_2}=0,6.22,4=13,44\left(l\right)\)
c) \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,6}{1}\) => CuO hết, H2 dư
PTHH: CuO + H2 --to--> Cu + H2O
0,1------------>0,1
=> mchất rắn = 0,1.64 = 6,4 (g)
\(n_{HCl}=0,4.1=0,4\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,2 0,4 0,2
\(m_{Zn}=0,2.65=13\left(g\right)\\
V_{H_2}=0,2.22,4=4,48\left(l\right)\\
n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\\
pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\
LTL:\dfrac{0,25}{1}>\dfrac{0,2}{1}\)
=> CuO dư
\(n_{CuO\left(p\text{ư}\right)}=n_{Cu}=n_{H_2}=0,2\left(mol\right)\\
X=\left\{{}\begin{matrix}m_{CuO\left(d\right)}=\left(0,25-0,2\right).80=4\left(g\right)\\m_{Cu}=0,2.64=12,8\left(g\right)\end{matrix}\right.=4+12,8=16,8\left(g\right)\)
\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.1.................................0.1\)
\(Đặt:n_{CuO\left(pư\right)}=x\left(mol\right)\)
\(CuO+H_2\underrightarrow{t^0}Cu+H_2O\)
\(x............x\)
\(m_{cr}=6-80x+64x=5.2\left(g\right)\)
\(\Rightarrow x=0.05\)
\(H\%=\dfrac{0.05}{0.075}\cdot100\%=66.67\%\)
Câu 3:
c, Từ phần trên, có nH2 = nFe = 0,1 (mol)
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{3}\), ta được Fe2O3 dư.
Theo PT: \(n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{15}\left(mol\right)\Rightarrow m_{Fe}=\dfrac{1}{15}.56=\dfrac{56}{15}\left(g\right)\)
a) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: `Fe + 2HCl -> FeCl_2 + H_2`
0,1-->0,2----->0,1------>0,1
`=> m_{FeCl_2} = 0,1.127 = 12,7 (g)`
b) `V_{H_2} = 0,1.22,4 = 2,24 (l)`
c) `n_{Fe_2O_3} = (16)/(160) = 0,1 (mol)`
PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\)
Xét tỉ lệ: \(0,1>\dfrac{0,1}{3}\Rightarrow\) Fe2O3
Theo PT: \(n_{Fe}=\dfrac{2}{3}.n_{H_2}=\dfrac{1}{15}\left(mol\right)\)
\(\Rightarrow m_{Fe}=\dfrac{1}{15}.56=\dfrac{56}{15}\left(g\right)\)
\(a,n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ PTHH:Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\left(1\right)\\ Theo.pt\left(1\right):n_{H_2}=n_{MgSO_4}=n_{Mg}=0,1\left(mol\right)\\ V_{H_2}=0,1.22,4=2,24\left(l\right)\\ m_{MgSO_4}=0,1.120=12\left(g\right)\\ b,PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\left(2\right)\\ Theo.pt\left(2\right):n_{Cu}=n_{H_2}=0,1\left(mol\right)\\ m_{Cu}=0,1.64=6,4\left(g\right)\)
\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,3 0,3 0,3
\(m_{MgCl_2}=0,3.95=28,5g\\ V_{H_2}=0,3.22,4=6,72l\\ n_{CuO}=\dfrac{3}{80}=0,0375\left(mol\right)\\ pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ LTL:\dfrac{0,0375}{1}>\dfrac{0,3}{1}\)
=>Hidro dư
\(n_{Cu}=n_{CuO}=0,0375\left(mol\right)\\ m_{Cu}=0,0375.64=2,4\left(g\right)\)