a)

Làm giấy quỳ tím hóa xanh là dung dịch Bazo

\(CaO+H_2O->Ca\left(OH\right)_2\)

b)

Làm giấy quỳ tím hóa đỏ là dung dịch axit

\(SO_3+H_2O->H_2SO_4\)

c)

\(Zn+2HCl->ZnCl_2+H_2\)

d)

\(2KClO_3->2KCl+3O_2\)

\(n_{Fe_2o_3}=\dfrac{36}{160}=0,2\left(m\right)\)

\(PTHH:Fe_2O_3+3H_2\xrightarrow[]{}2Fe+3H_2O\)

tỉ lệ          :1             3           2         3

số mol     :0,2          0,6        0,4       0,6

\(m_{Fe}=0,4.56=22,4\left(g\right)\)

b)\(V_{H_2}=0,6.22,4=13,44\left(l\right)\)

\(m_{HCl}=100.7,3\%=7,3\left(g\right)\Rightarrow n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

PT: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{BaCl_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\)

\(\Rightarrow C\%_{BaCl_2}=\dfrac{0,1.208}{100+100}.100\%=10,4\%\)

Câu 9:

Ta có: \(m_{BaCl_2}=200.26\%=52\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{52}{208}=0,25\left(mol\right)\)

\(n_{H_2SO_4}=0,3.1=0,3\left(mol\right)\)

PT: \(BaCl_2+H_2SO_4\rightarrow2HCl+BaSO_{4\downarrow}\)

Xét tỉ lệ: \(\dfrac{0,25}{1}< \dfrac{0,3}{1}\), ta được H₂SO₄ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{HCl}=2n_{BaCl_2}=0,5\left(mol\right)\\n_{BaSO_4}=n_{H_2SO_4\left(pư\right)}=n_{BaCl_2}=0,25\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,05\left(mol\right)\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)

\(m_{BaSO_4}=0,25.233=58,25\left(g\right)\)

\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)

Câu 10:

\(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{1}{0,234}\approx4,2735\left(M\right)\)

Câu 11:

\(n_{HCl\left(2M\right)}=0,7.2=1,4\left(mol\right)\)

\(n_{HCl\left(3M\right)}=1,1.3=3,3\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{1,4+3,3}{0,7+1,1}\approx2,611\)

a, \(n_{CO_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PT: \(MgO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2O\)

\(MgCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O\)

Theo PT: \(n_{MgCO_3}=n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{0,1.84}{10,4}.100\%\approx80,77\%\\\%m_{MgO}\approx19,23\%\end{matrix}\right.\)

b, \(n_{MgO}=\dfrac{10,4-0,1.84}{40}=0,05\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=2n_{MgO}+2n_{MgCO_3}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)