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

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

        1          2             1           1

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

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

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

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

b) \(n_{H2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)

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

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

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

c) \(n_{HCl}=\dfrac{36,5}{36,5}=1\left(mol\right)\)

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

        1          2              1        1

      0,2         1              0,2

Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{1}{2}\)

           ⇒ Zn phản ứng hết , Hcl dư

          ⇒ Tính toán dựa vào số mol của Zn

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

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

\(n_{HCl\left(dư\right)}=1-\left(0,2.2\right)=0,6\left(mol\right)\)

⇒ \(m_{HCl\left(dư\right)}=0,6.36,5=14,6\left(g\right)\)

 Chúc bạn học tốt

a, Ta có: n_Zn=\(\dfrac{13}{65}\)=0,2 mol

Zn + 2HCl ---> ZnCl₂ + H₂

Ta có: n_Zn=\(\dfrac{1}{2}\)n_HCl => n_HCl=0,1 mol

=> m_HCl=0,1.36,5=3,65 g

=> a%=\(\dfrac{3,65.100}{100}\)=3,65%

b, Ta có: n_Zn=n_ZnCl2 = n_H2= 0,2 mol

=> V_H2=0,2.22,4=4,48 l

=> m_ZnCl2=0,2.136=27,2 g

c, Zn + 2HCl ---> ZnCl₂ + H₂

Ta có: n_HCl=\(\dfrac{36.5}{36.5}\)=1 mol

Ta có: \(\dfrac{n_{HCl}}{n_{Zn}}=\dfrac{1}{0,2}\) => HCl dư tính theo Zn

Ta có: n_Zn=n_ZnCl2 = n_H2= 0,2 mol

=> V_H2=0,2.22,4=4,48 l

=> m_ZnCl2=0,2.136=27,2 g

\(a.n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\\ m_{ZnCl_2}=0,2.136=27,2\left(g\right)\\ V_{H_2}=0,2.22,4=4,48\left(l\right)\\ b.n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\\ CuO+H_2-^{t^o}\rightarrow Cu+H_2O\\ LTL:\dfrac{0,15}{1}< \dfrac{0,2}{1}\\ \Rightarrow H_2dưsauphảnứng\\ n_{Cu}=n_{H_2\left(pứ\right)}=n_{H_2O}=n_{CuO}=0,15\left(mol\right)\\ m_{Cu}=0,15.64=9,6\left(g\right)\\ m_{H_2\left(dư\right)}=\left(0,2-0,15\right).2=0,1\left(g\right)\\ m_{H_2O}=0,15.18=2,7\left(g\right)\)

nZn = 13/65 = 0,2 (mol)

PTHH: Zn + 2HCl -> ZnCl2 + H2

Mol: 0,2 ---> 0,4 ---> 0,2 ---> 0,2

mZnCl2 = 0,2 . 136 = 27,2 (g)

VH2 = 0,2 . 22,4 = 4,48 (l)

nCuO = 12/80 = 0,15 (mol)

PTHH: CuO + H2 -> (t°) Cu + H2O

LTL: 0,15 < 0,2 => H2 dư

nCuO (p/ư) = nCu = nH2O = nCuO = 0,15 (mol)

mCu = 0,15 . 64 = 9,6 (g)

mH2O = 0,15 . 18 = 2,7 (g)

mCuO (dư) = (0,2 - 0,15) . 80 = 4 (g)

Zn+2HCl->ZnCl₂+H₂

0,2-------------0,2----0,2

n Zn=\(\dfrac{13}{65}\)=0,2 mol

=>m _ZnCl2=0,2.136=27,2g

=>V_H2=0,2.22,4=4,48l

c)H₂+CuO-to>Cu+H₂O

           0,15-----0,15---0.15

n _CuO=\(\dfrac{12}{80}\)=0,15 mol

=>H2 dư

=>m _Cu=0,15.64=9,6g

=>m _H2O=0,15.18=2,7g

=>m _{CuO dư}=0,05.80=4g

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

PTHH :

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

0,3      0,6         0,3         0,3 

\(a,V_{H_2}=n.22,4=0,3.22,4=6,72\left(l\right)\)

\(m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)

\(b,V_{ddHCl}=\dfrac{n}{C_M}=\dfrac{0,6}{2}=0,3\left(l\right)\)

\(c,Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)

     0,1         0,3 

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

Ta có : 

\(\dfrac{0,1}{1}=\dfrac{0,3}{3}\)

nên không chất nào dư 

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

\(Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\uparrow\)

0,1         →          0,1 → 0,1

a) \(V_{H_2}=22,4\cdot0,1=2,24\left(l\right)\)

b) \(m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\)

c) \(Zn+H_2SO_4\xrightarrow[]{}ZnSO_4+H_2\uparrow\)

bđ: 0,1 → 0,4

pư: 0,1 → 0,1

\(\Rightarrow H_2SO_4\text{ dư}\)

\(\Rightarrow n_{H_2SO_4\text{ dư}}=0,4-0,1=0,3\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\text{ dư}}=0,3\cdot98=29,4\left(g\right)\)

\(n_{Zn}=\dfrac{32,5}{65}=0,5mol\)

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

0,5      1             0,5         0,5

\(V_{H_2}=0,5\cdot22,4=11,2l\)

\(m_{ZnCl_2}=0,5\cdot136=68g\)

ớ ớ ớ ớ ớ

Zn + 2HCl ---> ZnCl2 + H2 

0.2-→0.4------→0.2--→0.2 (mol)

nZn = 11,2\56 = 0.2(mol)

mZnCl2 = n*M = 0.2*127 = 25.4(g)

VH2(đktc) = n*22.4 = 0.2*22.4 = 4.48(l)

mHCl = n*M = 0.4*36.5 = 14.6(g)

C% =14,6\146*100% = 10(%)

\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.n_{Zn}=\dfrac{11,2}{65}=0,17\left(mol\right)\\ TheoPT:n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,17\left(mol\right)\\ m_{ZnCl_2}=0,17.136=23,12\left(g\right)\\ V_{H_2}=0,17.22,4=3,808\left(l\right)\\ c.n_{HCl}=2n_{Zn}=0,34\left(mol\right)\\ C\%_{HCl}=\dfrac{0,34.36,5}{146}.100=8,5\%\)

PTHH: 2Al + 6HCl -> 2AlCl3 + 3H2 

nAl = 5.4/27= 0.2mol

Theo ptpư : nH2 = 3/2 nAl = 3/2.0.2 = 0.3 mol 

=> VH2 = 0.3.22.4= 6.72 (lít)

HCl dư , tính theo Al 

nAlCl3 = nAl = 0.2 mol 

mAlCl3 = 0.2×133.5= 26.7 g 

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