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a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Gọi: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\) ⇒ 65x + 27y = 17,05 (1)
Ta có: \(n_{H_2}=\dfrac{9,52}{22,4}=0,425\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y=0,425\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,2\left(mol\right)\\n_{Al}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,2.65=13\left(g\right)\\m_{Al}=0,15.27=4,05\left(g\right)\end{matrix}\right.\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\n_{AlCl_3}=n_{Al}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnCl_2}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\end{matrix}\right.\)
c, Ta có: m dd HCl = 1,05.500 = 525 (g)
m dd sau pư = mhh + m dd HCl - mH2 = 541,2 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,2.136}{541,2}.100\%\approx5,03\%\\C\%_{AlCl_3}=\dfrac{0,15.133,5}{541,2}.100\%\approx3,7\%\end{matrix}\right.\)
a, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PT: \(M+H_2SO_4\rightarrow MSO_4+H_2\)
Theo PT: \(n_M=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow M_M=\dfrac{11,2}{0,2}=56\left(g/mol\right)\)
→ M là Fe.
b, Theo PT: \(n_{FeSO_4}=n_{H_2SO_4\left(pư\right)}=n_{H_2}=0,2\left(mol\right)\)
⇒ nH2SO4 dư = 0,5.1 - 0,2 = 0,3 (mol)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0,3}{0,5}=0,6\left(M\right)\\C_{M_{FeSO_4}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\end{matrix}\right.\)
c, Ta có: \(n_{FeSO_4.7H_2O}=n_{FeSO_4}=0,2\left(mol\right)\)
\(\Rightarrow m_{FeSO_4.7H_2O}=0,2.278=55,6\left(g\right)\)
Bài 1:
nH2SO4 bđ = 0,5 . 0,2 = 0,1 mol
nH2 = \(\dfrac{1,792}{22,4}=0,08\left(mol\right)\)
Pt: Zn + H2SO4 --> ZnSO4 + H2
............0,08 mol<-0,08 mol<-0,08 mol
Theo pt: nH2SO4 pứ = nH2 = 0,08 mol < 0,1 mol
=> HCl dư
CM H2SO4 dư = \(\dfrac{\left(0,1-0,08\right)}{0,2}=0,1M\)
CM ZnSO4 = \(\dfrac{0,08}{0,2}=0,4M\)
Bài 2:
nCuO = \(\dfrac{3,2}{80}=0,04\left(mol\right)\)
mH2SO4 = \(\dfrac{150\times32,666}{100}=49\left(g\right)\)
nH2SO4 = \(\dfrac{49}{98}=0,5\left(mol\right)\)
Pt: CuO + H2SO4 --> CuSO4 + H2O
0,04 mol->0,04 mol->0,04 mol
Xét tỉ lệ mol giữa CuO và H2SO4:
\(\dfrac{0,04}{1}< \dfrac{0,5}{1}\)
Vậy H2SO4 dư
mH2SO4 dư = (0,5 - 0,04) . 98 = 45,08 (g)
mCuSO4 = 0,04 . 160 = 6,4 (g)
mdd sau pứ = mCuO + mdd H2SO4 = 3,2 + 150 = 153,2 (g)
C% dd CuSO4 = \(\dfrac{6,4}{153,2}.100\%=4,177\%\)
C% dd H2SO4 dư = \(\dfrac{45,08}{153,2}.100\%=29,425\%\)
a, Gọi \(m_{NaCl\left(thêm\right)}=a\left(g\right)\)
\(m_{NaCl\left(bđ\right)}=5\%.100=5\left(g\right)\\ \Rightarrow C\%_{NaCl}=\dfrac{5+a}{100+a}.100\%=5,5\%\\ \Leftrightarrow a=0,53\left(g\right)\)
b, \(m_{NaCl}=58,5.5,5\%=3,2175\left(g\right)\\ n_{NaCl}=\dfrac{3,2175}{58,5}=0,055\left(mol\right)\)
PTHH: NaCl + AgNO3 ---> AgCl↓ + NaNO3
0,055-->0,055------>0,055---->0,055
\(m_{AgCl}=0,055.143,5=7,8925\left(g\right)\\ m_{ddY}=58,5+200-7,8925=250,6075\left(g\right)\\ \Rightarrow C\%_{NaNO_3}=\dfrac{0,055.85}{250,6075}.100\%=1,87\%\)
a)
C% CuSO4 = 16/(16 + 184) .100% = 8%
b)
n NaOH = 20/40 = 0,5(mol)
CM NaOH = 0,5/4 = 0,125M
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
\(n_{HCl}=\dfrac{400.14,6\%}{36,5}=1,6\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
Xét tỉ lệ: \(\dfrac{0,4}{2}< \dfrac{1,6}{6}\) => Al hết, HCl dư
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,4------------->0,4---->0,6
mdd sau pư = 10,8 + 400 - 0,6.2 = 409,6 (g)\(C\%_{AlCl_3}=\dfrac{0,4.133,5}{409,6}.100\%=13,037\%\)
