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1)
- Xét phần 1:
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,2<-------------------0,2
=> nFe = 0,2 (mol)
- Xét phần 2:
\(n_{SO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: 2Fe + 6H2SO4 --> Fe2(SO4)3 + 3SO2 + 6H2O
0,2-->0,6-------->0,1--------->0,3
Cu + 2H2SO4 --> CuSO4 + SO2 + 2H2O
0,3<----0,6<------0,3<-----0,3
=> nCu = 0,3 (mol)
m = 2.(0,2.56 + 0,3.64) = 60,8 (g)
2)
\(m_{H_2SO_4\left(bđ\right)}=\dfrac{200.98}{100}=196\left(g\right)\)
=> \(m_{H_2SO_4\left(sau.pư\right)}=196-98\left(0,6+0,6\right)=78,4\left(g\right)\)
mdd sau pư = \(\dfrac{60,8}{2}+200-0,6.64=192\left(g\right)\)
\(\left\{{}\begin{matrix}C\%_{\left(Fe_2\left(SO_4\right)_3\right)}=\dfrac{0,1.400}{192}.100\%=20,83\%\\C\%_{\left(CuSO_4\right)}=\dfrac{0,3.160}{192}.100\%=25\%\\C\%_{\left(H_2SO_4.dư\right)}=\dfrac{78,4}{192}.100\%=40,83\%\end{matrix}\right.\)
\(n_{CaCO_3}=\dfrac{40}{100}=0,4\left(mol\right)\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ n_{CO_2}=n_{CaCO_3}=0,4\left(mol\right)\\ \Rightarrow n_{NaOH}=0,4.2=0,8\left(mol\right)\\ m_{NaOH}=0,8.40=32\left(g\right)\\ \Rightarrow m_{ddNaOH}=\dfrac{32.100}{25}=128\left(g\right)\\ \Rightarrow V_{ddNaOH}=\dfrac{m_{ddNaOH}}{D_{ddNaOH}}=\dfrac{128}{1,28}=100\left(ml\right)=0,1\left(l\right)\)
- PT: a, \(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\) (1)
\(MnO_2+4HCl_đ\underrightarrow{t^o}MnCl_2+Cl_2+2H_2O\) (2)
- Ta có: \(n_{HCl\left(1\right)}=n_{HCl\left(2\right)}=0,2.2=0,4\left(mol\right)\)
Theo PT (1): \(n_{Cl_2}=\dfrac{5}{16}n_{HCl\left(1\right)}=0,125\left(mol\right)\Rightarrow V_1=0,125.22,4=2,8\left(l\right)\)
(2): \(n_{Cl_2\left(2\right)}=\dfrac{1}{4}n_{HCl\left(2\right)}=0,1\left(mol\right)\Rightarrow V_2=0,1.22,4=2,24\left(l\right)\)
\(a,\) Đặt hóa trị của M là \(x(x>0)\)
\(n_{H_2}=\dfrac{0,672}{22,4}=0,03(mol)\\ PTHH:2M+2xHCl\to 2MCl_x+xH_2\\ \Rightarrow n_{M}=\dfrac{0,03}{x}.2=\dfrac{0,06}{x}(mol)\\ \Rightarrow M_M=\dfrac{0,72}{\dfrac{0,06}{x}}=12x\)
Thay \(x=2\Rightarrow M_M=24(g/mol)\)
Vậy M là magie (Mg)
\(b,n_{HCl}=0,5.0,2=0,1(mol)\)
Vì \(\dfrac{n_{HCl}}{2}>\dfrac{n_{H_2}}{1}\) nên \(HCl\) dư
\(\Rightarrow n_{MgCl_2}=n_{H_2}=0,03(mol)\\ \Rightarrow C_{M_{MgCl_2}}=\dfrac{0,03}{0,2}=0,15M\)
Chọn D
Gọi N 2 C O 3 (x mol) và N a H C O 3 ( y mol)
Nhỏ từ từ HCl vào hỗn hợp xảy ra pư theo thứ tự:
\(a,PTHH:X+2HCl\to XCl_2+H_2\\ \Rightarrow n_{X}=n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ \Rightarrow M_X=\dfrac{9,75}{0,15}=65(g/mol)(Zn)\\ b,n_{HCl}=2.0,2=0,4(mol)\)
Vì \(\dfrac{n_{H_2}}{1}<\dfrac{n_{HCl}}{2}\) nên \(HCl\) dư
\(\Rightarrow n_{ZnCl_2}=n_{H_2}=0,15(mol)\\ \Rightarrow m_{ZnCl_2}=136.0,15=20,4(g)\\ C_{M_{ZnCl_2}}=\dfrac{0,15}{0,2}=0,75M\)
\(HCl+NaHCO_3\rightarrow NaCl+CO_2+H_2O\\ N_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ m_{HCl}=n_{CO_2}=0,2\left(mol\right)\\ V_{ddHCl}=\dfrac{112}{1,12}=100\left(ml\right)=0,1\left(l\right)\\ \Rightarrow C_{MddHCl}=\dfrac{0,2}{0,1}=2\left(M\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ C\%_{ddHCl}=\dfrac{7,3}{112}.100\approx6,518\%\)
\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: NaHCO3 + HCl --> NaCl + CO2 + H2O
0,2<------------0,2
=> \(C\%\left(HCl\right)=\dfrac{0,2.36,5}{112}.100\%=6,518\%\)
\(V_{ddHCl}=\dfrac{112}{1,12}=100\left(ml\right)=0,1\left(l\right)\)
=> \(C_M\left(HCl\right)=\dfrac{0,2}{0,1}=2M\)