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

a, PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

______0,2_____0,2_________0,2 (mol)

b, \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,2}{0,5}=0,4M\)

c, \(m_{CaCO_3}=0,2.100=20\left(g\right)\)

Bạn tham khảo nhé!

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

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

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

Rainbow

1.

\(Mg+H_2SO_4-->MgSO_4+H_2\)

\(MgSO_4+2NaOH-->Mg\left(OH\right)_2+Na_2SO_4\)

\(Mg\left(OH\right)_2-->MgO+H_2O\)

\(MgO+2HCl-->MgCl_2+H_2O\)

2. \(n_{HCl}=\dfrac{200\cdot18,25}{100.36,5}=1\left(mol\right)\)

a) PTHH : \(Ca\left(OH\right)_2+2HCl-->CaCl_2+2H_2O\)

Theo pthh : \(n_{Ca\left(OH\right)2}=\dfrac{1}{2}n_{HCl}=0,5\left(mol\right)\)

=> \(m_{ddCa\left(OH\right)2}=\dfrac{74.0,5\cdot100}{10}=370\left(g\right)\)

b) Theo pthh : \(n_{CaCl_2}=n_{Ca\left(OH\right)2}=0,5\left(mol\right)\)

=> \(m_{CaCl2}=55,5\left(g\right)\)

Áp dụng DDLBTKL :

m_{(dd Ca(oh)2)} + m_(ddHCl) = m_(ddCaCl2)

=> 370 + 200 = 570 g

=> \(C\%CaCl_2=\dfrac{55,5}{570}\cdot100\%=9,74\%\)

a)

$Fe +H_2SO_4 \to FeSO_4 + H_2$

$FeSO_4 + 2KOH \to Fe(OH)_2 + K_2SO_4$
$4Fe(OH)_2 + O_2 \xrightarrow{t^o} 2Fe_2O_3 + 4H_2O$

$n_{Fe_2O_3} = \dfrac{20}{160} = 0,125(mol)$

Theo PTHH : $n_{Fe} = 2n_{Fe_2O_3} = 0,25(mol)$
$m_{Fe} = 0,25.56 = 14(gam)$

b)

$n_{H_2} = n_{Fe} = 0,25(mol)$
$V_{H_2} = 0,25.22,4 = 5,6(lít)$
c)

$n_{H_2SO_4} = n_{Fe} = 0,25(mol)$
$V_{dd\ H_2SO_4} = \dfrac{0,25}{1} = 0,25(lít) = 250(ml)$

\(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ FeSO_4+2KOH\rightarrow Fe\left(OH\right)_2+K_2SO_4\\4 Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)

\(a.n_{Fe_2O_3}=\dfrac{20}{160}=0,125\left(mol\right)\\ n_{H_2}=n_{H_2SO_4}=n_{Fe}=n_{FeSO_4}=n_{Fe\left(OH\right)_2}=\dfrac{4}{2}.0,125=0,25\left(mol\right)\\ m_{Fe}=0,25.56=14\left(g\right)\\ b.V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\\ c.V_{ddH_2SO_4}=\dfrac{0,25}{1}=0,25\left(l\right)=250\left(ml\right)\)