Nồng độ %? hoặc 500 ml dd HCl

\(n_{CaCO_3}=\dfrac{30}{100}=0,3mol\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ n_{HCl}=0,3.2=0,6mol\\ \left[{}\begin{matrix}C_{M_{HCl}}\\C_{\%HCl}=\dfrac{0,6.36,5}{500}\cdot100=4,38\%\end{matrix}\right.=\dfrac{0,6}{0,5}=1,2M}\)

\(SO_3+H_2O\rightarrow H_2SO_4\\ H_2SO_4+Na_2SO_3\rightarrow Na_2SO_4+SO_2+H_2O\\ SO_2+Na_2O\rightarrow Na_2SO_3\\ Na_2SO_3+2HCl\rightarrow2NaCl+SO_2+H_2O\)

1, SO₃+H₂O->H₂SO₄

2, H₂SO₄+Na₂SO₃->Na₂SO₄+H₂O+SO₂

3, SO₂+2NaOH->Na₂SO₃+H₂O

4, Na₂SO₃+H₂SO₄->Na₂SO₄+H₂O+SO₂

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.

A: MgO, CuO

B: MgCl₂, CuCl₂

C: Mg(OH)₂, Cu(OH)₂

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

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

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

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

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

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

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

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

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

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

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