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a) Gọi số mol Zn, Fe là a, b (mol)
=> 65a + 56b = 8,56 (1)
\(n_{H_2}=\dfrac{3,136}{22,4}=0,14\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
a--->2a-------->a----->a
Fe + 2HCl --> FeCl2 + H2
b----->2b------->b------>b
=> a + b = 0,14 (2)
(1)(2) => a = 0,08; b = 0,06
=> \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,08.65}{8,56}.100\%=60,748\%\\\%m_{Fe}=\dfrac{0,06.56}{8,56}.100\%=39,252\%\end{matrix}\right.\)
b)
nKOH = 0,2.0,1 = 0,02 (mol)
PTHH: KOH + HCl --> KCl + H2O
0,02-->0,02
=> nHCl = 0,02 + 2a + 2b = 0,3 (mol)
=> \(C_{M\left(HCl\right)}=xM=\dfrac{0,3}{0,15}=2M\)
c) m = 0,08.136 + 0,06.127 = 18,5(g)
a. Gọi \(a,b\) lần lượt là số mol của \(Mg,Fe\) có trong hỗn hợp ban đầu.
\(\Rightarrow m_{hh}=24a+56b=22,8\left(g\right)\left(1\right)\)
\(n_{SO_2}=\frac{15,68}{22,4}=0,7\left(mol\right)\)
Các quá trình oxi hóa và khử:
\(Mg\rightarrow Mg^{2+}+2e\)
\(a----->2a\)
\(Fe\rightarrow Fe^{3+}+3e\)
\(b----->3b\)
\(S^{+6}+2e\rightarrow S^{+4}\)
\(0,7-1,4-0,7\)
Áp dung định luật bào toàn electron, ta có: \(2a+3b=1,4\left(mol\right)\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\) ta được hệ: \(\left\{{}\begin{matrix}24a+56b=22,8\\2a+3b=1,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,25\left(mol\right)\\b=0,3\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Mg}=0,25.24=6\left(g\right)\Rightarrow\%m_{Mg}=\frac{6}{22,8}=.100\%=26,32\%\)
\(\Rightarrow\%m_{Fe}=100\%-26,32\%=73,68\%\)
b. Từ câu a, ta được: \(n_{H_2SO_4\left(p.ư\right)}=n_{S^{+6}}=0,7\left(mol\right)\)
\(n_{NaOH}=0,4.3=1,2\left(mol\right)\)
PTHH: \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=\frac{1,2}{2}=0,6\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(bđ\right)}=0,7+0,6=1,3\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(bđ\right)}=1,3.98=127,4\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4\left(bđ\right)}=\frac{127,4.100}{98}=130\left(g\right)\)
c. Gọi \(x\) là số mol của \(Cu_2S\) \(\rightarrow n_{FeS_2}=2x\left(mol\right)\)
PTHH:
\(2FeS_2+14H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+14H_2O+15SO_2\)
\(2x---->14x-->x\)
\(Cu_2S+6H_2SO_4\rightarrow2CuSO_4+5SO_2+6H_2O\)
\(x---->6x-->2x\)
\(\Rightarrow14x+6x=1,3\left(mol\right)\Leftrightarrow x=0,065\left(mol\right)\)
\(\Rightarrow n_{Fe_2\left(SO_4\right)_3}=x=0,065\left(mol\right);n_{CuSO_4}=2x=0,13\left(mol\right)\)
\(\Rightarrow m_{Fe_2\left(SO_4\right)_3}=0,065.400=26\left(g\right);m_{CuSO_4}=0,13.160=20,8\left(g\right)\)
\(\Rightarrow m\)muối\(=26+20,8=46,8\left(g\right)\)
a) \(n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,05<-----------0,05---->0,075
=> \(\%Al=\dfrac{0,05.27}{14,15}.100\%=9,54\%\)
=> \(\%Cu=\dfrac{14,15-0,05.27}{14,15}.100\%=90,46\%\)
b) \(V_{H_2}=0,075.22,4=1,68\left(l\right)\)
c) \(n_{Cu}=\dfrac{14,15-0,05.27}{64}=0,2\left(mol\right)\)
PTHH: 4Al + 3O2 --to--> 2Al2O3
0,05->0,0375
2Cu + O2 --to--> 2CuO
0,2-->0,1
=> \(V_{O_2}=\left(0,1+0,0375\right).22,4=3,08\left(l\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ m_{AlCl_3}=6,675\left(mol\right)\\ n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\\ \Rightarrow n_{Al}=n_{AlCl_3}=0,05\left(mol\right)\\ \Rightarrow m_A=0,05.27=1,35\left(g\right);m_{Cu}=14,15-1,35=12,8\left(g\right)\\ \%m_{Cu}=\dfrac{12,8}{14,15}.100\approx90,459\%\\ \Rightarrow\%m_{Al}\approx9,541\%\\ b,n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ \Rightarrow V=V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ 2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{O_2}=\dfrac{3}{4}.n_{Al}+\dfrac{1}{2}.n_{Cu}=\dfrac{3}{4}.0,05+\dfrac{1}{2}.0,2=0,0875\left(mol\right)\)
\(\Rightarrow V_{O_2\left(đktc\right)}=0,0875.22,4=1,96\left(l\right)\)