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a, \(Na_2O+H_2O\rightarrow2NaOH\)
Ta có: \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\)
\(\Rightarrow CM_{NaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\)
b, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\\ n_{Mg}=n_{H_2SO_4}=n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ a,m_{Mg}=0,15.24=3,6\left(g\right)\\ b,C_{MddH_2SO_4}=\dfrac{0,15}{0,2}=0,75\left(M\right)\)
a, \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)
b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
\(n_{NaOH}=2n_{Na_2O}=0,2\left(mol\right)\Rightarrow C_{M_{NaOH}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)
\(a,n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH:
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,15 0,3 0,15 0,15
\(m_{Fe}=0,15.56=8,4\left(g\right)\)
\(a,n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
PTHH :
\(Na_2O+H_2O\rightarrow2NaOH\)
0,1 0,1 0,2
\(C_{M\left(A\right)}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)
a) Na2O +H2O-->2NaOH (1)
2NaOH +CuSO4 -->Na2SO4+ Cu(OH)2 (2)
Cu(OH)2 -to-> CuO +H2O (3)
b) mNa2O=8.100/100=8(g)
=>nNa2O=8/62=0,13(mol)
theo(2) :nCu(OH)2=1/2nNaOH=0,065(mol)
theo(3):nCuO=nCu(OH)2=0,065(mol)
=>mCuO=0,065.80=5,2(g)
c) CuO +2HCl-->CuCl2+H2O (4)
theo (4) : nHCl=2nCuO=0,13(mol)
mddHCl 25%=0,13.36,5.100250,13.36,5.10025=18,98(g)
\(n_{Na_2O}=\dfrac{6,2}{62}=0,1 \left(mol\right)\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
0,1 -----------------> 0,1
\(CM_{base}=CM_{NaOH}=\dfrac{0,1}{0,2}=0,5M\)
b
\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
0,05 <------ 0,1
\(V_{H_2SO_4}=\dfrac{0,05}{0,2}=0,25\left(l\right)\Rightarrow V_{dd.H_2SO_4}=\dfrac{0,25.100}{20}=1,25\left(l\right)\)
a,\(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
PTHH: Na2O + H2O → 2NaOH
Mol: 0,1 0,2
\(\Rightarrow C_{M_{ddA}}=\dfrac{0,2}{0,1}=2M\)
b,mddKOH = 6,2+100.1=106,2 (g)
\(\Rightarrow D_{ddKOH}=\dfrac{106,2}{100}=1,062\left(g/cm^3\right)\)
c,mKOH = 0,2.56 = 11,2 (g)
\(C\%_{ddKOH}=\dfrac{11,2.100\%}{106,2}=10,55\%\)
a, \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
b, \(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)
Theo PT: \(n_{CuO}=n_{CuCl_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,05.80=4\left(g\right)\)
c, \(C_{M_{CuCl_2}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
\(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)
PTHH :
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
0,05 0,1 0,05
\(b,m_{CuO}=0,05.80=4\left(g\right)\)
\(c,C_{M\left(CuCl_2\right)}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
nNa2O=0.1(mol)
pthh:Na2O+H2O->2NaOH
theo pthh:nH2O=nNa2O->nH2O=0.1(mol)->mH2O=0.1*18=1.8(g)
b)Theo pthh:nNaOH=2 nNa2O
->nNaOH=0.1*2=0.2(mol)
CM=0.2:0.2=1(M)
Khk bt là gì bạn? =)) câu b mình không tìm thấy lỗi sai.bạn chỉ cho mình với?