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

1, SO₃+H₂O->H₂SO₄

2, H₂SO₄+Na₂SO₃->Na₂SO₄+H₂O+SO₂

3, SO₂+2NaOH->Na₂SO₃+H₂O

4, Na₂SO₃+H₂SO₄->Na₂SO₄+H₂O+SO₂

\(\left(1\right)3Fe+2O_2\xrightarrow[]{t^0}Fe_3O_4\\ \left(2\right)Fe_3O_4+8HCl\rightarrow2FeCl_3+FeCl_2+4H_2O\\ \left(3\right)FeCl_2+Ag_2SO_4\rightarrow FeSO_4+2AgCl\\ \left(4\right)FeSO_4+Ba\left(NO_3\right)_2\rightarrow Fe\left(NO_3\right)_2+BaSO_4\\ \left(5\right)Fe\left(NO_3\right)_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaNO_3\\ \left(6\right)Fe\left(OH\right)_2\xrightarrow[]{t^0}FeO+H_2O\)

cho tui hỏi 98 chia 16 bằng bao nhiêu vậy tui đang gấp

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

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

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

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

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

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

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

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