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\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\\ n_{H_2}=n_{Fe}=0,25\left(mol\right)\\ V_{H_2}=0,25.22,4=5,6\left(l\right)\\ n_{HCl}=2n_{Fe}=0,5\left(mol\right)\\ m_{HCl}=0,5.36,5=18,25\left(g\right)\)
a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2--->0,3-------->0,1----------->0,3
b) `V_{H_2} = 0,3.22,4 = 6,72 (l)`
c) `m_{H_2SO_4} = 0,3.98 = 29,4 9g)`
d) \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
Xét tỉ lệ: 0,2 < 0,3 => H2 dư
\(n_{Fe}=\dfrac{28}{56}=0,5mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,5 1 0,5
\(V_{H_2}=0,5\cdot22,4=11,2l\)
\(m_{HCl}=1\cdot36,5=36,5g\)
a.b.\(n_{Mg}=\dfrac{3,6}{24}=0,15mol\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,15 0,3 0,15 ( mol )
\(V_{H_2}=0,15.22,4=3,36l\)
\(m_{HCl}=0,3.36,5=10,95g\)
c.\(n_{H_2}=0,15.60\%=0,09mol\)
\(Ag_2O+H_2\rightarrow\left(t^o\right)2Ag+H_2O\)
0,09 0,18 ( mol )
\(m_{Ag}=0,18.108=19,44g\)
\(a.n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ TheoPT:n_{H_2}=n_{Mg}=0,15\left(mol\right)\\ \Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)\\ b.TheoPT:n_{HCl}=2n_{Mg}=0,3\left(mol\right)\\ \Rightarrow m_{Mg}=0,3.36,5=10,95\left(g\right)\\ c.n_{H_2\left(pứ\right)}=0,15.60\%=0,054\left(g\right)\\ H_2+Ag_2O-^{t^o}\rightarrow2Ag+H_2O\\ n_{Ag}=2n_{H_2}=0,108\left(mol\right)\\ \Rightarrow m_{Ag}=0,108.108=11,664\left(g\right)\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1\left(mol\right)\\ m_{muối}=m_{ZnCl_2}=0,1.136=13,6\left(g\right)\\ V_{khí\left(đktc\right)}=V_{H_2\left(đkc\right)}=0,1.24,79=2,479\left(l\right)\\ c,n_{CuO}=\dfrac{7,6}{80}=0,095\left(mol\right)\\ PTHH:CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Vì:\dfrac{0,095}{1}< \dfrac{0,1}{1}\Rightarrow H_2dư\\ n_{Cu}=n_{CuO}=0,095\left(mol\right)\\ m_{Cu}=0,095.64=6,08\left(g\right)\)
nFe=0,1 mol
Fe +2HCl=>FeCl2+H2
0,1 mol--0,2 mol -0,1 mol
VH2=0,1.22,4=2,24 lít
=>mHCl=0,2.36,5=7,3g
2H2+O2-to->2H2O
=>nO2 =1,12\22,4=0,05 mol
=> pứ hết >mH2O=0,1.18=1,8g
bài hỏi kẽm mà bn ghi sắt ???