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![](https://rs.olm.vn/images/avt/0.png?1311)
1.
Al2O3 + 2NaOH -> 2NaAlO2 + H2O (1)
nNaAlO2=0,225(mol)
Từ 1:
nNaOH=nNaAlO2=0,225(mol)
nal2O3=\(\dfrac{1}{2}\)nNaAlO2=0,1125(mol)
V dd NaOH=0,225:5=0,045(lít)
mAl2O3=0,1125.102=11,475(g)
mquặng=11,475.110%=12,6225(g)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(m_{dd_{HCl\left(10\%\right)}}=150\cdot1.206=180.9\left(g\right)\)
\(n_{HCl}=\dfrac{180.9\cdot10\%}{36.5}\approx0.5\left(mol\right)\)
\(n_{HCl\left(2M\right)}=0.25\cdot2=0.5\left(mol\right)\)
\(n_{HCl}=0.5+0.5=1\left(mol\right)\)
\(V_{dd_{HCl}}=150+250=400\left(ml\right)=0.4\left(l\right)\)
\(C_{M_{HCl}}=\dfrac{1}{0.4}=2.5\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mdd HCl 10% = 150.1,047 = 157,05 (g)
=> \(n_{HCl\left(dd.HCl.10\%\right)}=\dfrac{157,05.10\%}{36,5}=\dfrac{3141}{7300}\left(mol\right)\)
nHCl(dd HCl 2M) = 0,25.2 = 0,5 (mol)
=> \(C_{M\left(A\right)}=\dfrac{\dfrac{3141}{7300}+0,5}{0,15+0,25}=\dfrac{6791}{2920}M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(a.n_{NaOH\left(tổng\right)}=0,05.1+0,2.0,2=0,09\left(mol\right)\\ V_{ddNaOH\left(tổng\right)}=50+200=250\left(ml\right)=0,25\left(l\right)\\ C_{MddNaOH\left(cuối\right)}=\dfrac{0,09}{0,25}=0,36\left(M\right)\\ b.n_{HCl}=0,5.0,02=0,01\left(mol\right)\\ n_{H_2SO_4}=0,08.0,2=0,016\left(mol\right)\\ V_{ddsau}=20+80=100\left(ml\right)=0,1\left(l\right)\\ C_{MddH_2SO_4}=\dfrac{0,016}{0,1}=0,16\left(M\right)\\ C_{MddHCl}=\dfrac{0,01}{0,1}=0,1\left(M\right)\)
Bài 2:
\(a.m_{H_2SO_4}=29,4.10\%=2,94\left(g\right)\\ b.n_{H_2SO_4}=\dfrac{2,94}{98}=0,03\left(mol\right)\\ n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,01}{1}< \dfrac{0,03}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(dư\right)}=0,03-0,01=0,02\left(mol\right)\\ m_{H_2SO_4\left(dư\right)}=0,02.98=1,96\left(g\right)\\ n_{H_2}=n_{Fe}=0,01\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,01.22,4=0,224\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right);n_{HCl}=2.0,2=0,4\left(mol\right)\\ m_{ZnCl_2}=136.0,2=27,2\left(g\right);C_{MddHCl}=\dfrac{0,4}{0,1}=4\left(M\right)\\ b,Zn+CuSO_4\rightarrow ZnSO_4+Cu\\ n_{CuSO_4}=\dfrac{20.10\%}{160}=0,0125\left(mol\right);n_{Zn}=0,2\left(mol\right)\\ Vì:\dfrac{0,0125}{1}< \dfrac{0,2}{1}\Rightarrow Zn.dư\\ n_{Zn\left(p.ứ\right)}=n_{ZnSO_4}=n_{CuSO_4}=0,0125\left(mol\right)\\m_{Zn\left(p.ứ\right)}=0,0125.65=0,8125\left(g\right)\\ m_{ddsau}=m_{Zn\left(p.ứ\right)}+m_{ddCuSO_4}=0,8125+20=20,8125\left(g\right)\\ C\%_{ddZnSO_4}=\dfrac{0,0125.161}{20,8125}.100\approx9,67\%\)
a) \(m_{ddNaOH}=\frac{20}{10\%}=200\left(g\right)\)
b) \(m_{H_2SO_4}=0,1\times98=9,8\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\frac{9,8}{10\%}=98\left(g\right)\)
c) \(m_{ddH_2SO_4}=300\times0,8=240\left(g\right)\)
d) \(m_{ddCa\left(OH\right)_2}=\frac{7,4}{10\%}=74\left(g\right)\)
e) \(m_{HCl}=0,2\times36,5=7,3\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\frac{7,3}{15\%}=48,67\left(g\right)\)
a)
mddNaOH = 20*100/10=200g
b)
mH2SO4 =0.1*98 = 9.8 g
mddH2SO4 = 9.8*100/10=98g
c)
mddHCl = 300*0.8=240g
d)
mddCa(OH)2 = 7.4*100/10=74g
e)
mHCl = 0.2*36.5=7.3g
mddHCl = 7.3*100/15= 48.67g