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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)PTHH: Ca + 2H2O\(\rightarrow\) Ca(OH)2 + H2 (1)
CaO + H2O \(\rightarrow\)Ca(OH)2 (2)
b) nH2= \(\dfrac{2,24}{22,4}\)=0,1 mol
Theo PT1: nCa=nH2= 0,1 mol
=> mCa=0,1x40=4 g
=>%mCa=\(\dfrac{4}{9,6}\)x100%=41,67%
=>%mCaO=100%-41,67%=58,33%
c) mCaO=9,6-4=5,6g
nCaO=\(\dfrac{5,6}{56}\)=0,1 mol
Theo PT1và PT2 có: nCa+nCaO=nCa(OH)2(PT1) + nCa(OH)2(PT2)
=> nCa(OH)2(thu đc)=0,1+0,1=0,2 mol
=> mCa(OH)2=0,2 x 74 = 14,8 g
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
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
=> \(m_{Ca\left(OH\right)_2}=\left(0,1+0,1\right).74=14,8\left(g\right)\)
Bài 14:
a) \(n_{H_2}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\)
PTHH: Ca + 2H2O --> Ca(OH)2 + H2
0,5<--------------0,5<----0,5
=> mCa = 0,5.40 = 20 (g)
=> \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{20}{34}.100\%=58,82\%\\\%m_{CaO}=100\%-58,82\%=41,18\%\end{matrix}\right.\)
b) b phải là khối lượng bazo thu được chứ nhỉ..., sao tính đc m dung dịch
\(n_{CaO}=\dfrac{34-20}{56}=0,25\left(mol\right)\)
PTHH: CaO + H2O --> Ca(OH)2
0,25---------->0,25
=> mCa(OH)2 = (0,5 + 0,25).74 = 55,5 (g)
\(n_{H_2}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\)
\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)
0,5 0,5 0,5 ( mol )
( \(CaO+H_2O\) không giải phóng \(H_2\) )
\(m_{Ca}=0,5.40=20g\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{20}{34}.100=58,82\%\\\%m_{CaO}=100\%-58,82\%=41,18\%\end{matrix}\right.\)
\(n_{CaO}=\dfrac{34-20}{56}=0,25\left(mol\right)\)
\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
0,25 0,25 ( mol )
\(m_{Ca\left(OH\right)_2}=\left(0,5+0,25\right).74=55,5g\)
a) Ca + 2H2O → Ca(OH)2 + H2↑ (1)
CaO + H2O → Ca(OH)2 (2)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
b) Theo Pt1: \(n_{Ca}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Ca}=0,1\times40=4\left(g\right)\)
\(\Rightarrow m_{CaO}=9,6-4=5,6\left(g\right)\)
\(\Rightarrow\%Ca=\dfrac{4}{9,6}\times100\%=41,67\%\)
\(\%CaO=\dfrac{5,6}{9,6}\times100\%=58,33\%\)
b) Theo PT1: \(n_{Ca\left(OH\right)_2}=n_{H_2}=0,1\left(mol\right)\)
\(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PT2: \(n_{Ca\left(OH\right)_2}=n_{CaO}=0,1\left(mol\right)\)
\(\Rightarrow\Sigma n_{Ca\left(OH\right)_2}=0,1+0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{Ca\left(OH\right)_2}=0,2\times74=14,8\left(g\right)\)