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

PTHH: CO₂ + Ca(OH)₂ → CaCO₃ ↓  +  H₂O

Mol:     0,25      0,25            0,25 

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

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

c, 

PTHH: Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

Mol:       0,25           0,5

\(m_{ddHCl}=\dfrac{0,5.36,5.100}{20}=91,25\left(g\right)\)

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

a) \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

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

Theo PTHH: \(n_{Ca\left(OH\right)_2}=n_{CO_2}=0,1\left(mol\right)\)

\(V_{Ca\left(OH\right)_2}=200ml=0,2l\)

\(\Rightarrow C_{MCa\left(OH\right)_2}=\dfrac{n_{Ca\left(OH\right)_2}}{V_{Ca\left(OH\right)_2}}=\dfrac{0,1}{0,2}=0,5M\)

b) Theo PTHH có: \(n_{CaCO_3}=n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=n_{CaCO_3}.M_{CaCO_3}=0,1.74=7,4\left(g\right)\)

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!

a) $CuSO_4 + 2NaOH \to Cu(OH)_2 + Na_2SO_4$

b) $n_{Cu(OH)_2} = n_{CuSO_4} = \dfrac{16}{160} = 0,1(mol)$
$m_{Cu(OH)_2} = 0,1.98 = 9,8(gam)$

c) $n_{NaOH} = 2n_{CuSO_4} = 0,2(mol) \Rightarrow C_{M_{NaOH}} = \dfrac{0,2}{0,1} = 2M$

PTHH: \(CuSO_4+2NaOH\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

Ta có: \(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{Cu\left(OH\right)_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2\left(M\right)\\m_{Cu\left(OH\right)_2}=0,1\cdot98=9,8\left(g\right)\end{matrix}\right.\)

a, \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

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

\(n_{Ca\left(OH\right)_2}=n_{CO_2}=0,25\left(mol\right)\Rightarrow C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,2}=1,25\left(M\right)\)

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

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

\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{15\%}\approx121,67\left(g\right)\)

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol\\ a)CO_2+Ca\left(OH\right)_2\rightarrow Ca\left(OH\right)_2+H_2O\)

    0,25       0,25                0,25

\(b)C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,2}=1,25M\\ c)2HCl+Ca\left(OH\right)_2\rightarrow CaCl_2+2H_2O\\ n_{HCl}=2n_{Ca\left(OH\right)_2}=2.0,25=0,5mol\\ m_{ddHCl}=\dfrac{0,5.36,5}{15\%}\cdot100\%\approx121,67g\)

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

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

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

Đổi 1 bar = 0,98692 atm

a+b) \(n_{CO_2}=\dfrac{PV}{RT}=\dfrac{0,98692\cdot6,2}{0,082\cdot\left(25+273\right)}=0,25\left(mol\right)=n_{CaCO_3}=n_{Ca\left(OH\right)_2}\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\\m_{CaCO_3}=0,25\cdot100=25\left(g\right)\end{matrix}\right.\)

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

Theo PTHH: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\) \(\Rightarrow m_{ddHCl}=\dfrac{0,5\cdot36,5}{20\%}=91,25\left(g\right)\)

1

a, CO2+Ca(OH)2--->CaCO3+H2O

nCO2=0,25 mol

=>nCa(OH)2=nCO2=0,25 mol

\(C_M=\frac{0,25}{0,1}=2,5\)M

b,

nCaCO3=nCO2=0,25mol

=>mCaCO3=0,25.100=25 g

c,

Ca(OH)2+2HCl--->CaCl2+H2O

nHCl=2.nCa(OH)2=0,25.2=0,5 mol

=>mHCl=0,5.36,5=18,25 g

=>mddHCl=18,25.100/20=91,25 g

2

Fe+H2SO4---->FeSO4+H2

nFe=0,56/56=0,01 mol

nFeSO4=nFe=0,01 mol

=>nFeSO4=0,01.152=1,52 g

nH2=nH2SO4=0,01 mol

=>VH2=0,01.22,4=0.224 lít

b,CM H2SO4=0,01/0,2=0,05M