a) M_X=\(\frac{40}{71,43\%}\)= 56 g/mol

Dễ thấy chỉ có 1 mol Canxi và 1 mol Oxi

=> CTHH: CaO

CaO + H₂O -> Ca(OH)₂

n _Ca(OH)2=\(\frac{14,8}{74}\)=0,2 mol

n_CaO=0,2 mol

m_CaO=0,2.56=11,2 g

c)n_H2O=0,2 mol

m_H2O=0,2.18=3,6 g

V_H2O=3,6/1=3,6 ml

Bài 1:

a) \(2Mg+O_2\underrightarrow{t^0}2MgO\)

b) \(n_{MgO}=\frac{8}{40}=0,2\left(mol\right)\)

Theo PTHH: \(n_{MgO}:n_{Mg}=2:2=1\)

\(\Rightarrow n_{Mg}=n_{MgO}=0,2\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,2.24=4,8\left(g\right)\)

Theo PTHH: \(n_{MgO}:n_{O_2}=2:1\)

\(\Rightarrow n_{O_2}=n_{Mg}.\frac{1}{2}=0,2.\frac{1}{2}=0,1\left(mol\right)\)

\(\Rightarrow V_{O_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)

Bài 2 nhé

Gọi CTTQ: Ca_xO_y

\(\dfrac{40x}{16y}=\dfrac{71,43}{28,57}\)

=> \(\dfrac{x}{y}=\dfrac{71,43.16}{28,57.40}=\dfrac{1}{1}\)

=> CTHH: CaO

a) nCa(OH)2 = 0,2 mol

Pt: CaO + H₂O --> Ca(OH)2

.....0,2<----0,2<--------0,2........(mol)

mCaO cần = 0,2 . 56 = 11,2 (g)

b) V = \(\dfrac{0,2.18}{1}=3,6\left(ml\right)\)

Gọi CTHH của hợp chất X là Ca_xO_y

Ta có: \(40x\div16y=71,43\div28,57\)

\(\Rightarrow x\div y=\dfrac{71,43}{40}\div\dfrac{28,57}{16}\)

\(\Rightarrow x\div y=1\div1\)

Vậy CTHH của hợp chất X là CaO

CaO + H₂O → Ca(OH)₂

a) \(n_{Ca\left(OH\right)_2}=\dfrac{14,8}{74}=0,2\left(mol\right)\)

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

\(\Rightarrow m_{CaO}=0,2\times56=11,2\left(g\right)\)

b) \(n_{CaO}=\dfrac{56}{56}=1\left(mol\right)\)

Theo PT: \(n_{H_2O}=n_{CaO}=1\left(mol\right)\)

\(\Rightarrow m_{H_2O}=1\times18=18\left(g\right)\)

\(\Rightarrow V_{H_2O}=\dfrac{18}{1}=18\left(ml\right)\)

a. PTHH: CaC₂ + 2H₂O ===> C₂H₂ + Ca(OH)₂

b. Áp dụng định luật bảo toàn khối lượng, ta có:

\(m_{CaC2}+m_{H2O}=m_{C_2H_2}+m_{Ca\left(OH\right)2}\)

\(\Leftrightarrow m_{H2O}=m_{C_2H_2}+m_{Ca\left(OH\right)2}-m_{CaC2}\)

\(\Leftrightarrow m_{H2O}=13+37-41=9\left(gam\right)\)

Áp dụng công thức \(V=\frac{m}{D}\)

=> Thể tích nước cần dùng là:

\(V_{H2O}=\frac{m_{H2O}}{D_{H2O}}=\frac{9}{1}=9\left(ml\right)\)

\(a,\) Cacbon Canxi + Nước → Axetylen + Canxi Hidroxit

\(b,\text{Bảo toàn KL: }m_{CaC_2}+m_{H_2O}=m_{C_2H_2}+m_{Ca\left(OH\right)_2}\\ \Rightarrow m_{H_2O}=13+37-41=9\left(g\right)\\ \Rightarrow V_{H_2O}=\dfrac{9}{1}=9\left(ml\right)\)

nCO2= 0,15(mol)

PTHH: CO2 + Ca(OH)2 -> CaCO3 + H2O

m(đá vôi)= mCaCO3= 0,15.100=15(g)

a, CO₂ + Ca(OH)₂ \(\rightarrow\) CaCO₃ + H₂O

b, \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15mol\)

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

0,15                        0,15                        ( mol )

\(\Rightarrow m_{CaCO_3}=0,15.100=15g\)

\(PTHH:CO_2+Ca\left(OH\right)_2 \rightarrow CaCO_3 +H_2 O\\ n_{CO_2}=0,15\left(mol\right)\\ m_{đá.vôi}=m_{CaCO_3}=0,15.100=15\left(g\right)\). 

a. PTHH:

\(Ca+2H_2O--->Ca\left(OH\right)_2+H_2\left(1\right)\)

\(CaO+H_2O--->Ca\left(OH\right)_2\left(2\right)\)

b. Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT₍₁₎: \(n_{Ca}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Ca}=0,1.40=4\left(g\right)\)

\(\Rightarrow\%_{m_{Ca}}=\dfrac{4}{9,6}.100\%=41,7\%\)

\(\%_{m_{CaO}}=100\%-41,7\%=58,3\%\)

c. Ta có: \(n_{CaO}=\dfrac{9,6-4}{56}=0,1\left(mol\right)\)

Ta có: \(n_{hh}=0,1+0,1=0,2\left(mol\right)\)

Theo PT_(1,2): \(n_{Ca\left(OH\right)_2}=n_{hh}=0,2\left(mol\right)\)

\(\Rightarrow m_{Ca\left(OH\right)_2}=0,2.74=14,8\left(g\right)\)

a)  Ca+2H₂O→Ca(OH)₂+H₂

      0,1                  0,1       0,1         mol

     CaO+H₂O→Ca(OH)₂

        0,1               0,1                mol

b)nH₂=2,24/22,4=0,1 mol

mCa=0,1.40=4 g

%mCa=4/9,6 .100%=41,6 %

%mCaO=100%-41,6%=58,4%

c)mCaO=9,6-4=5,6 g

nCaO=5,6/56= 0,1 mol

mCa(OH)₂=0,1.148=14,8 g

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