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cứu với

a, \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

Ta có: \(n_{Al_2O_3}=\dfrac{15,3}{102}=0,15\left(mol\right)\)

b, \(n_{AlCl_3}=2n_{Al_2O_3}=0,3\left(mol\right)\Rightarrow m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\)

c, \(n_{HCl}=6n_{Al_2O_3}=0,9\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,9.36,5}{14,6\%}=225\left(g\right)\)

d, m _{dd sau pư} = 15,3 + 225 = 240,3 (g)

\(\Rightarrow C\%_{AlCl_3}=\dfrac{40,05}{240,3}.100\%\approx16,67\%\)

e, n_H2SO4 = 0,3.1 = 0,3 (mol)

PT: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,3}{3}\), ta được Al₂O₃ dư.

→ Al₂O₃ không tan hết.

Ta có: n_{CuCl2 (4M)} = 0,22.4 = 0,88 (mol)

Gọi: V_{CuCl2 (1,2M)} = a (l)

⇒ n_{CuCl2 (1,2M)} = 1,2a (mol)

\(\Rightarrow\dfrac{0,88+1,2a}{0,22+a}=2\)

⇒ a = 0,55 (l) = 550 (ml)

A: MgO, CuO

B: MgCl₂, CuCl₂

C: Mg(OH)₂, Cu(OH)₂

PT: \(2Mg+O_2\underrightarrow{t^o}2MgO\)

\(2Cu+O_2\underrightarrow{t^o}2CuO\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)

\(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_{2\downarrow}\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

\(P_{Fe}=V_{Fe}.D_{Fe}=100.18000=1800000\left(N\right)\)

\(n_{FeO}=\dfrac{2,16}{72}=0,03\left(mol\right)\\ n_{HCl}=0,2.0,4=0,08\left(mol\right)\\ a,PTHH:FeO+2HCl\rightarrow FeCl_2+H_2O\\ Vì:\dfrac{0,03}{1}< \dfrac{0,08}{2}\Rightarrow HCldư\\ n_{FeCl_2}=n_{FeO}=0,03\left(mol\right)\\ m_{FeCl_2}=127.0,03=3,81\left(g\right)\\ n_{HCl\left(Dư\right)}=0,08-2.0,03=0,02\left(mol\right)\\ V_{ddsau}=V_{ddHCl}=0,4\left(l\right)\\ C_{MddHCl\left(dư\right)}=\dfrac{0,02}{0,4}=0,05\left(M\right);C_{MddFeCl_2}=\dfrac{0,03}{0,4}=0,075\left(M\right)\)

\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\\a, CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4}=n_{H_2SO_4}=n_{CuO}=0,05\left(MOL\right)\\ b,m_{CuSO_4}=0,05.160=8\left(g\right)\\ c,V_{ddH_2SO_4}=\dfrac{0,05}{0,5}=0,1\left(l\right)\\ d,V_{ddCuSO_4}=V_{ddH_2SO_4}=0,1\left(l\right)\\ C_{MddCuSO_4}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)

\(a)4Al+3O_2\xrightarrow[]{t^0}2Al_2O_3\\ b)n_{Al_2O_3}=\dfrac{20,4}{102}=0,2mol\\ n_{Al}=\dfrac{0,2.4}{2}=0,4mol\\ m_{Al}=0,4.27=10,8g\)

c) bạn xem lại đề

\(a)7437ml=7,437l\\ n_{N_2}=\dfrac{7,437}{24,79}=0,3mol\\ m_{N_2}=0,3.28=8,4g\\ b)n_{Cl_2}=\dfrac{2,479}{24,79}=0,1mol\\ m_{Cl_2}=0,1.71=7,1g\)

b, 14,9 g

Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

Theo ĐLBT KL: m_Zn + m_H2SO4 = m_ZnSO4 + m_H2

⇒ m_H2SO4 = 9 + 0,1.2 - 6,5 = 2,7 (g)