Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

a. PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂

b. Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)

=> \(V_{H_2}=0,1.22,4=2,24\left(lít\right)\)

c. Theo PT: \(n_{H_2SO_4}=n_{Zn}=0,1\left(mol\right)\)

=> \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)

Ta có: \(C_{M_{H_2SO_4}}=\dfrac{9,8}{m_{dd_{H_2SO_4}}}.100\%=25\%\)

=> \(m_{dd_{H_2SO_4}}=39,2\left(g\right)\)

Ta có: \(m_{H_2}=0,1.2=0,2\left(g\right)\)

=> \(m_{dd_{ZnSO_4}}=6,5+39,2-0,2=45,5\left(g\right)\)

Theo PT: \(n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)

=> \(m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)

=> \(C_{\%_{ZnSO_4}}=\dfrac{16,1}{45,5}.100\%=35,4\%\)

a, \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

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

Theo PT: \(n_{H_2SO_4}=n_{MgSO_4}=n_{H_2}=n_{Mg}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.24,79=2,479\left(l\right)\)

c, \(m_{ddH_2SO_4}=\dfrac{0,1.98}{19,6\%}=50\left(g\right)\)

d, Ta có: m _{dd sau pư} = 2,4 + 50 - 0,1.2 = 52,2 (g)

\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,1.120}{52,2}.100\%\approx22,99\%\)

Câu 1 :

a)  PTHH : 

 \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) (1) 

  \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)(2)

b) Ta có : \(n_{Zn}=\frac{3,5}{65}\approx0,054\left(mol\right)\)

Theo phương trình hóa học (1) :

\(n_{H_2}=n_{Zn}\approx0,054\left(mol\right)\)

\(\Rightarrow V_{H_2}\approx0,054\cdot22,4=1,2096\left(l\right)\)

c) Theo phương trình hóa học (2), ta có:

\(n_{Cu}=n_{H_2}\approx0,054\left(mol\right)\)

\(\Rightarrow m_{Cu}\approx0,054\cdot64=3,456\left(g\right)\)

Bài 2:

a) Ta có : \(n_{Al}=\frac{2,7}{27}=0,1\left(mol\right)\); \(n_{HCl}=\frac{200\cdot7,3}{100\cdot36,5}=0,4\left(mol\right)\)

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Theo phương trình hóa học : \(n_{H_2}=\frac{3}{2}n_{Al}=\frac{3}{2}.0,1=0,15\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)

b) Theo phương trình hóa học , ta có : \(n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,1\cdot133,5=13,35\left(g\right)\)

Lại có: \(m_{H_2}=0,15\cdot2=0,3\left(g\right)\)

Áp dụng định luật bảo toàn khối lượng, ta có:

\(m_{Al}+m_{dd_{HCl}}=m_{dd_{AlCl_3}}+m_{H_2}\)

\(\Leftrightarrow2,7+200=m_{dd_{AlCl_3}}+0,3\)

\(\Leftrightarrow m_{dd_{AlCl_3}}=202,4\left(g\right)\)

Vậy \(C\%_{dd_{AlCl_3}}=\frac{13,35}{202,4}\cdot100\%\approx6,6\%\)

\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)

\(n_{HCl}=\dfrac{100\cdot14.6\%}{36.5}=0.4\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(1........2\)

\(0.1......0.4\)

\(LTL:\dfrac{0.1}{1}< \dfrac{0.4}{2}\Rightarrow HCldư\)

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{\text{dung dịch sau phản ứng}}=6.5+100-0.1\cdot2=106.3\left(g\right)\)

\(C\%ZnCl_2=\dfrac{0.1\cdot136}{106.3}\cdot100\%=12.79\%\)

\(C\%HCl\left(dư\right)=\dfrac{\left(0.4-0.2\right)\cdot36.5}{106.3}\cdot100\%=6.87\%\%\)

Bài 4 : 

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

            1          2              1          1

          0,2        0,4           0,2        0,2

b) \(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(C_{ddHCl}=\dfrac{14,6.100}{100}=14,6\)⁰/₀

d) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)

\(m_{ddspu}=13+100-\left(0,2.2\right)=112,6\left(g\right)\)

\(C_{ZnCl2}=\dfrac{27,2.100}{112,6}=24,16\)⁰/₀

 Chúc bạn học tốt

Bài 3 : 

\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)

a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2|\)

            1            1                1           1

          0,5          0,5            0,5          0,5

b) \(n_{H2}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,5.22,4=11,2\left(l\right)\)

c) \(n_{H2SO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

⇒ \(m_{H2SO4}=0,5.98=49\left(g\right)\)

\(C_{ddH2SO4}=\dfrac{49.100}{200}=24,5\)⁰/₀

d) \(n_{MgSO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

⇒ \(m_{MgSO4}=0,5.120=60\left(g\right)\)

\(m_{ddspu}=12+200-\left(0,5.2\right)=211\left(g\right)\)

\(C_{MgSO4}=\dfrac{60.100}{211}=28,44\)⁰/₀

 Chúc bạn học tốt

Mg+H2SO4->MgSO4+H2

0,5----0,5--------0,5--------0,5

n Mg=12\24=0,5 mol

=>VH2=0,5.22,4=11,2l

=>m H2SO4=0,5.98=49 g

=>C% =49\200.100=24,5g

=>m dd=12+200-0,5.2=211g

=>C%=0,4.120\211.100=22,74885%

\(n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\\ m_{HCl}=\dfrac{100.18,25}{100}=18,25\left(g\right)\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)  
            0,125                            0,125 (mol ) 
\(\Rightarrow V_{H_2}=0,125.22,4=2,8\left(l\right)\\ \)   
\(C\%=\dfrac{8,125}{8,125+18,25}.100\%=30,8\%\)
 

Bài 18:

Ta có: \(n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\)

\(m_{HCl}=100.18,25\%=18,25\left(g\right)\Rightarrow n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, Xét tỉ lệ: \(\dfrac{0,125}{1}< \dfrac{0,5}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Zn}=0,125\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,125.22,4=2,8\left(g\right)\)

\(m_{H_2}=0,125.2=0,25\left(g\right)\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,125\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Zn}=0,25\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,25\left(mol\right)\)

Có: m _{dd sau pư} = 8,125 + 100 - 0,25 = 107,875 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,125.136}{107,875}.100\%\approx15,76\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,25.36,5}{107,875}.100\%\approx8,46\%\end{matrix}\right.\)

Bạn tham khảo nhé!

`a)PTHH:`

`Zn + 2HCl -> ZnCl_2 + H_2↑`

`0,125`  `0,25`    `0,125`    `0,125`      `(mol)`

`b)n_[Zn] = [ 8,125 ] / 65 = 0,125 (mol)`

`n_[HCl] = [ [ 18,25 ] / 100 . 100 ] / [ 36,5 ] = 0,5 (mol)`

Ta có: `[ 0,125 ] / 1 < [ 0,5 ] / 2`

    `-> Zn` hết, `HCl` dư

 `=> V_[H_2] = 0,125 . 22,4 = 2,8 (l)`

 `=> m_[H_2] = 0,125 . 2 = 0,25 (g)`

`c)m_\text{dd sau p/ứ} = 8,125 + 100 - 0,25 = 107,875 (g)`

`=> C%_[ZnCl_2] = [ 0,125 . 136 ] / [ 107,875 ] . 100 ~~ 15,76%`

`=> C%_[HCl(dư)] = [ ( 0,5 - 0,25 ) . 36,5 ] / [ 107,875 ] . 100 ~~ 8,46%`

n Zn= 19,5/65=0,3 (mol).

PTPƯ: Zn(0.3) +   HCl(0.6) ---->  ZnCl2(0.3) + H2(0,3)

mHCl=0,6.36.5=21.9(g)

a) C%HCl= 21.9/300.100%=7,3%

b) VH2=0,3.22,4=6,72(lít)

c) mH2=0,3.2=0,6(g)

mZnCl2=0,3.136=40,8(g)

mddZnCl2 =(19,5+300)-0,6=318,9(g)

C%=mZnCl2/mddZnCl2.100= 40,8/318,9.100=12,793%

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1\left(mol\right)\\ m_{muối}=m_{ZnCl_2}=0,1.136=13,6\left(g\right)\\ V_{khí\left(đktc\right)}=V_{H_2\left(đkc\right)}=0,1.24,79=2,479\left(l\right)\\ c,n_{CuO}=\dfrac{7,6}{80}=0,095\left(mol\right)\\ PTHH:CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Vì:\dfrac{0,095}{1}< \dfrac{0,1}{1}\Rightarrow H_2dư\\ n_{Cu}=n_{CuO}=0,095\left(mol\right)\\ m_{Cu}=0,095.64=6,08\left(g\right)\)

\(a) Fe + H_2SO_4 \to FeSO_4 +H_2\\ b) n_{H_2} = n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ c) m_{dd\ sau\ pư} = 5,6 + 224,6 - 0,1.2 = 230(gam)\\ n_{FeSO_4} = n_{Fe} = 0,1(mol)\\ C\%_{FeSO_4} = \dfrac{0,1.152}{230}.100\% = 6,61\%\\ d) ZnO + H_2 \xrightarrow{t^o} Zn + H_2O\\ n_{ZnO} = \dfrac{32,4}{81} = 0,4 > n_{H_2} = 0,1 \to ZnO\ dư\\ n_{ZnO\ pư} = n_{H_2} = 0,1(mol)\\ m_{ZnO\ dư} = 32,4 - 0,1.81 = 24,3(gam)\)