\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH:

Ca + 2H₂O ---> Ca(OH)₂ + H₂

0,1<-------------0,1<---------0,1

=> \(\left\{{}\begin{matrix}m_{Ca}=0,1.40=4\left(g\right)\\m_{CaO}=9,6-4=5,6\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{4}{9,6}.100\%=41,67\%\\\%m_{CaO}=100\%-41,67\%=58,33\%\end{matrix}\right.\)

\(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: CaO + H₂O ---> Ca(OH)₂

          0,1------------------>0,1

=> \(m_{Ca\left(OH\right)_2}=\left(0,1+0,1\right).74=14,8\left(g\right)\)

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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Chào em, em đăng bài nhầm môn, em đăng sang môn Hóa nha!

\(2Na+2H_2O\rightarrow2NaOH+H_2\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Na}=2.0,3=0,6\left(mol\right)\\ a,m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Na_2O}=26,2-13,8=12,4\left(g\right)\\b, n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ n_{NaOH\left(tổng\right)}=n_{Na}+2.n_{Na_2O}=0,6+\dfrac{12,4}{62}=0,8\left(mol\right)\\ m_{c.tan}=m_{NaOH}=0,8.40=32\left(g\right)\\ c,m_{ddNaOH}=m_{hh}+m_{H_2O}-m_{H_2}=26,2+200-0,3.2=225,6\left(g\right)\\ C\%_{ddNaOH}=\dfrac{32}{225,6}.100\approx14,185\%\)

Anh có làm rồi nha!

nH2 = 2.24/22.4 = 0.1 (mol) 

Zn + 2HCl => ZnCl2 + H2 

ZnO + 2HCl => ZnCl2 + H2O 

nZn = nH2 = 0.1 (mol) 

=> mZn = 6.5 g

mZnO = 14.6 - 6.5 = 8.1 (g) 

nZnO = 0.1 (mol)  

%Zn = 6.5/14.6 * 100% = 44.52%

%ZnO = 55.48%

nHCl = 0.1*2 + 0.1*2= 0.4 (mol) 

Vdd HCl = 0.4 / 0.5 = 0.8 l 

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

PT: \(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

Theo PT: \(n_{Ca}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Ca}=0,1.40=4\left(g\right)\)

Ta có: n_H2= 2,24/22,4 = 0,1 (mol)

 PTHH: Ca + 2H₂O → Ca(OH₂ + H₂↑ (1)

  (mol) 0,1        0,1 ←   0,1

          CaO + H₂O  Ca(OH)₂ (2)

     Tính phần trăm khối lượng:

Từ (1) → n_H2= n_Ca= 0,1 (mol) → m_Ca = 0,1 x 40 = 4 (gam)

Vậy %m_Ca = 4/9,6 x 100% = 41,667%;

     %m_CaO = 100% - 41,667% = 58,333%

     m_CaO = 9,6 – 4 = 5,6 (g) → n_CaO = 5,6/56 = 0,1 (mol)

Từ (1) và (2) → n_Ca(OH)2 =0,1+ 0,1 = 0,2 (mol)

          → m_Ca(OH)2= 0,2 x 74 = 14,8 (gam).   

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