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1-B

2: 1-B/2-B

3-D

4-A

5-B

6-C

\(CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O\)

Theo PTHH :

\(n_{CO_2} = n_{CaCO_3} = \dfrac{10}{100} = 0,1\ mol\)

\(n_{hỗn\ hợp\ khí} = \dfrac{11,2}{22,4} = 0,5(mol)\)

⇒ \(n_{CO} = 0,5 - 0,1 = 0,4(mol)\)

Suy ra : m_{hỗn hợp} = 0,1.44 + 0,4.28 = 15,6(gam)

Vậy :

\(\%m_{CO_2} = \dfrac{0,1.44}{15,6}.100\% = 28,21\%\\\%m_{CO} = 100\% -28,21\% = 71,79\%\)

\(a.n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ a.Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ 0,25.......0,25............0,25..........0,25\left(mol\right)\\ C_{MddCa\left(OH\right)_2}=\dfrac{0,25}{0,1}=2,5\left(M\right)\\ b.m_{\downarrow}=m_{CaCO_3}=100.0,25=25\left(g\right)\)

* Ta có: \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

a. PTHH: \(CO_2+Ca\left(OH\right)_2--->CaCO_3\downarrow+H_2O\)

b. Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CO_2}=0,25\left(mol\right)\)

Đổi 100ml = 0,1 lít

=> \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5M\)

* PTHH: X₂O₃ + 3H₂SO₄ ---> X₂(SO₄)₃ + 3H₂O

Đổi 600ml = 0,6 lít

Ta có: \(n_{H_2SO_4}=1.0,6=0,6\left(mol\right)\)

Theo PT: \(n_{X_2O_3}=\dfrac{1}{3}.n_{H_2SO_4}=\dfrac{1}{3}.0,6=0,2\left(mol\right)\)

=> \(M_{X_2O_3}=\dfrac{32}{0,2}=160\left(g\right)\)

Ta có: \(M_{X_2O_3}=NTK_X.2+16.3=160\left(g\right)\)

=> NTK_X = 56(đvC)

Vậy X là sắt (Fe)

=> CTHH là Fe₂O₃

a) $CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$

b) $n_{Ca(OH)_2} = n_{CO_2} = \dfrac{5,6}{22,4} = 0,25(mol)$

$\Rightarrow C_{M_{Ca(OH)_2}} = \dfrac{0,25}{0,1} = 2,5M$

c) $n_{CaCO_3} = n_{CO_2} = 0,25(mol)$

$\Rightarrow m_{CaCO_3} = 0,25.100 = 25(gam)$

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(n_{Ca\left(OH\right)_2}=0,1.0,5=0,05\left(mol\right)\)

Xét \(\dfrac{n_{CO_2}}{n_{Ca\left(OH\right)_2}}=\dfrac{0,2}{0,05}=4\)

=> Tạo ra muối HCO₃^-

PTHH: Ca(OH)₂ + 2CO₂ --> Ca(HCO₃)₂

______0,05------------------->0,05

=> m_Ca(HCO3)2 = 0,05.162=8,1(g)

muối không tan là CaCO₃

\(\Rightarrow\) n_CaCO3= \(\frac{40}{100}\)= 0,4( mol)

PTPU

Ca(OH)₂+ CO₂\(\rightarrow\) CaCO₃\(\downarrow\)+ H₂O

.0,4.............0,4........0,4................ mol

\(\Rightarrow\) C_{M Ca(OH)2}= \(\frac{0,4}{0,1}\)= 4M

theo PTPU có: n_CO2= n_CaCO3= 0,4( mol)

\(\Rightarrow\) %V_CO2= \(\frac{0,4.22,4}{11,2}\). 100%= 80%

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

Theo PT:  \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,25\left(mol\right)\)

a, \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)

b, \(m_{CaCO_3}=0,25.100=25\left(g\right)\)

c, \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

Theo PT: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{20\%}=91,25\left(g\right)\)