a) NaOH + HCl --> NaCl + H2O

b) Số mol NaOH là: n(NaOH) = 0,15mol

Theo pthh thì số mol NaCl là: n(NaCl) = 0,15mol

Khối lượng NaCl là: 0,15.58,5 = 8,775g

c)theo pthh thì số mol HCl là: 0,15mol

V(ddHCl)=100ml=0,1l

Nồng độ mol ddHCl là:

0,15/0,1=1,5M

\(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\a, CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ n_{NaOH}=2.0,05=0,1\left(mol\right)\\ b,C_{MddNaOH}=\dfrac{0,1}{0,1}=1\left(M\right)\\ c,n_{Na_2CO_3}=n_{CO_2}=0,05\left(mol\right)\\ m_{muối}=m_{Na_2CO_3}=0,05.106=5,3\left(g\right)\)

\(a.CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ b.n_{CO_2}=\dfrac{1,12}{22,4}=0,05mol\\ CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

0,05       0,1                  0,05

\(C_{M_{NaOH}}=\dfrac{0,1}{0,1}=1M\\ c.m_{Na_2CO_3}=0,05.106=5,3g\)

a, \(2KOH+MgSO_4\rightarrow K_2SO_4+Mg\left(OH\right)_2\)

b, Ta có: \(n_{KOH}=0,2.1=0,2\left(mol\right)\)

Theo PT: \(n_{MgSO_4}=n_{K_2SO_4}=n_{Mg\left(OH\right)_2}=\dfrac{1}{2}n_{KOH}=0,1\left(mol\right)\)

\(\Rightarrow m_{Mg\left(OH\right)_2}=0,1.58=5,8\left(g\right)\)

c, \(V_{MgSO_4}=\dfrac{0,1}{2}=0,05\left(l\right)\)

d, \(C_{M_{K_2SO_4}}=\dfrac{0,1}{0,2+0,05}=0,4\left(M\right)\)

\(a,n_{H_2SO_4}=0,5\cdot0,1=0,05\left(mol\right)\\ PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \Rightarrow n_{H_2}=n_{H_2SO_4}=0,05\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,05\cdot22,4=1,12\left(l\right)\\ b,n_{Al}=\dfrac{2}{3}n_{H_2SO_4}=\dfrac{1}{30}\left(mol\right)\\ \Rightarrow m_{Al}=\dfrac{1}{30}\cdot27=0,9\left(g\right)\\ c,n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2SO_4}\approx0,017\left(mol\right)\\ \Rightarrow C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,017}{0,1}\approx0,17M\)

PTHH: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

Ta có: \(n_{NaOH}=\dfrac{30\cdot20\%}{40}=0,15\left(mol\right)=n_{CH_3COONa}=n_{CH_3COOH}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CH_3COONa}=0,15\cdot82=12,3\left(g\right)\\C_{M_{CH_3COOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\end{matrix}\right.\)

a.b.\(n_{Zn}=\dfrac{1,95}{65}=0,03mol\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,03   0,06          0,03           0,03    ( mol )

\(V_{H_2}=0,03.22,4=0,672l\)

\(m_{ddHCl}=\dfrac{0,06.36,5}{7,3\%}=30g\)

c.Tên muối: Kẽm clorua

\(m_{ZnCl_2}=0,03.136=4,08g\)

\(m_{ddspứ}=30+1,95-0,03.2=31,89g\)

\(C\%_{ZnCl_2}=\dfrac{4,08}{31,89}.100\%=12,79\%\)

nhanh thế?

\(n_{Al}=\dfrac{21,6}{27}=0,8\left(mol\right)\)

Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O|\)

        2             3                  1                3

       0,8          1,2               0,4             1,2

a) \(n_{H2}=\dfrac{0,8.3}{2}=1,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=1,2.22,4=26,88\left(l\right)\)

b) \(n_{H2SO4}=\dfrac{0,8.3}{2}=1,2\left(mol\right)\)

⇒ \(m_{H2SO4}=1,2.98=117,6\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{117,6.100}{29,4}=400\left(g\right)\)

c) \(n_{Al2\left(SO4\right)3}=\dfrac{1,2.1}{3}=0,4\left(mol\right)\)

⇒ \(m_{Al2\left(SO4\right)3}=0,4.342=136,8\left(g\right)\)

\(m_{ddspu}=21,6+400-\left(1,2.2\right)=419,2\left(g\right)\)

\(C_{Al2\left(SO4\right)3}=\dfrac{136,8.100}{419,2}=32,63\)⁰/₀

 Chúc bạn học tốt

\(m_{H_2SO_4}=\dfrac{150.20}{100}=30\left(g\right)\)

a, \(n_{H_2SO_4}=\dfrac{30}{98}=\dfrac{15}{49}\left(mol\right)\)

PTHH :

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

          15/49      15/49       15/49

\(b,V_{H_2}=n.22,4=\dfrac{15}{49}.22,4=\dfrac{48}{7}\approx6.86\left(l\right)\)

\(c,m_{FeSO_4}=\dfrac{15}{49}.152\approx46,53\left(g\right)\)

Tên gọi : Sắt (II) Sunfat

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