a: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(n_{HCL}=\dfrac{10.95}{36.5}=0.3\left(mol\right)\)

\(n_{ZnCl_2}=n_{Zn}=\dfrac{13.6}{136}=0.1\left(mol\right)\)

Vì \(\dfrac{n_{Zn}}{1}< \dfrac{n_{HCl}}{2}\)

nên HCl dư

=>Tính theo mol của Zn

\(m_{Zn}=0.1\cdot65=6.5\left(g\right)\)

b: \(n_{H_2}=0.1\left(mol\right)\)

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

a) \(n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

\(n_{ZnCl_2}=\dfrac{13,6}{136}=0,1\left(mol\right)\)

PTHH : Zn + 2HCl -> ZnCl₂ + H₂

            0,1                     0,1 

Xét tỉ lệ \(\dfrac{0,3}{2}>\dfrac{0,1}{1}\) => HCl dư , ZnCl₂ đủ

\(m_{Zn}=0,1.65=6,5\left(g\right)\)

b. \(V_{Zn}=0,1.22,4=2,24\left(l\right)\)

\($a/$\\ Zn+2HCl \to ZnCl_2+H_2\\ b/\\ n_{Zn}=0,1(mol)\\ n_{HCl}=0,1.2=0,2(mol)\\ m_{ddHCl}=\frac{0,2.36,5.100}{7,3}=100(g)\\ C\%_{ZnCl_2}=\frac{0,1.136}{100+6,5-0,1.2}.100\%=12,8\%\)

\(a/\\ Zn+2HCl \to ZnCl_2+H_2\\ b/\\ n_{Zn}=\frac{6,5}{65}=0,1(mol)\\ n_{HCl}=0,2(mol)\\ m_{ddsaupu}=\frac{0,1.36,5.100}{7,3}+6,5-0,1.2=56,3(g)\\ C\%_{ZnCl_2}=\frac{0,1.136}{56,3}.100\%=24,16\%\)

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

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:     0,1    0,05       0,1

b, \(m_{HCl}=0,05.36,5=2,92\left(g\right)\)

  \(m_{ddHCl}=\dfrac{2,92}{7,3\%}=40\left(g\right)\)

  \(m_{ddsaupứ}=6,5+40-0,1.2=46,3\left(g\right)\)

\(\Rightarrow C\%_{ddZnCl_2}=\dfrac{0,1.136.100\%}{46,3}=29,4\%\)

PT: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\) (1)

\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\) (2)

\(Mg\left(OH\right)_2+2HCl\rightarrow MgCl_2+2H_2O\) (3) 

\(K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\) (4)

Ta có: \(m_{HCl}=10.95.20\%=2,19\left(g\right)\Rightarrow n_{HCl}=\dfrac{2,19}{36,5}=0,06\left(mol\right)\)

\(n_{CO_2}=\dfrac{0,224}{22,4}=0,01\left(mol\right)\)

Theo PT: \(n_{HCl\left(1\right)+\left(2\right)+\left(4\right)}=2n_{CO_2}=0,02\left(mol\right)\)

\(n_{H_2O\left(1\right)+\left(2\right)+\left(4\right)}=n_{CO_2}=0,01\left(mol\right)\)

\(\Rightarrow n_{HCl\left(3\right)}=0,06-0,02=0,04\left(mol\right)=n_{H_2O\left(3\right)}\)

⇒ n_H2O = 0,01 + 0,04 = 0,05 (mol)

Theo ĐLBT KL, có: m_hh + m_HCl = m _muối + m_CO2 + m_H2O

⇒ m = m _muối = 2,24 + 2,19 - 0,01.44 - 0,05.18 = 3,09 (g)

a)

$m_{Zn} + m_{HCl} = m_{ZnCl_2} + m_{H_2}$
$\Rightarrow m_{Zn} = 13,6 + 0,2 - 7,3 = 6,5(gam)$

b)

$Zn + 2HCl \to ZnCl_2 + H_2$

c)

Tỉ lệ số nguyên tử Zn : số phân tử HCl : số phân tử $ZnCl_2$ : số phân tử $H_2$ là 1 : 2 : 1 : 1

\(a,Phản.ứng.hóa.học:Zn+HCl->ZnCl_2+H_2\)

Theo định luật bảo toàn khối lượng:

 \(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ m_{Zn}+7,3g=13,6g+0,2g\\ m_{Zn}+7,3g=13,8g\\ m_{Zn}=13,8g-7,3g=6,5g\)

Vậy có 6,5g kẽm tham gia p/ứng.

\(b,PTHH:Zn+2HCl->ZnCl_2+H_2\)

c) Số nguyên tử Zn : Số phân tử HCl : Số phân tử ZnCl₂ : Số phân tử H₂ = 1:2:1:1

$a\big)$

$n_{Zn}=\dfrac{3,25}{65}=0,05(mol)$

$Zn+2HCl\to ZnCl_2+H_2$

Theo PT: $n_{ZnCl_2}=n_{Zn}=0,05(mol)$

$\to m_{ZnCl_2}=0,05.136=6,8(g)$

$b\big)$

Theo PT: $n_{HCl}=2n_{Zn}=0,1(mol)$

$\to V_{dd\,HCl}=\dfrac{0,1}{0,5}=0,2(l)=200(ml)$

`Zn + 2HCl -> ZnCl_2 + H_2`

`0,05`  `0,1`           `0,05`               `(mol)`

`n_[Zn]=[3,25]/65=0,05 (mol)`

`a)m_[ZnCl_2]=0,05.136=6,8(g)`

`b)V_[dd HCl]=[0,1]/[0,5]=0,2(M)`

a) lập PTHH của PƯ 

zn+2 hcl →zncl2 + h2

1mol   2mol   1mol   1mol 

0,1mol 0,2mol 

b) số mol của zn : nzn = 6,5: 65= 0,1mol

khối lượng Hcl : mhcl= 0,2 .  36,5=7,3(g)

c) c% hcl =(mhcl : mddhcl). 100=(7,3 .146): 100=10,658%

câu d) tương tự bạn tự làm nhé

Câu 1 : 

\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

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

\(m_{Zn}=0.1\cdot65=6.5\left(g\right)\)

Câu 2 : 

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(n_{HCl}=\dfrac{10.95}{36.5}=0.3\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

\(0.2...........0.3\)

\(LTL:\dfrac{0.2}{1}>\dfrac{0.3}{2}\Rightarrow Fedư\)

\(m_{Fe\left(dư\right)}=\left(0.2-0.15\right)\cdot56=2.8\left(g\right)\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

Câu 3 : 

Axit HCl em nhé !

\(n_{HCl}=\dfrac{9.125}{36.5}=0.25\left(mol\right)\)

\(R+2HCl\rightarrow RCl_2+H_2\)

\(0.125..0.25\)

\(M_R=\dfrac{7}{0.125}=56\left(\dfrac{g}{mol}\right)\)

\(R:Fe\left(Sắt\right)\)

nHCl=0,3(mol)

nZn=0,1(mol)

PTHH: Zn +2  HCl -> ZnCl2 + H2

ta có: 0,3/2 > 0,1/1

=> Zn hết, HCl dư, tính theo nZn.

=> nH2=nZn=0,1(mol)

=>V(H2,đktc)=0,1.22,4=2,24(l)

Chúc em học tốt!

$n_{Zn} = 0,1(mol) ; n_{HCl} = \dfrac{10,95}{36,5} = 0,3(mol)$

$Zn + 2HCl \to ZnCl_2 + H_2$
$n_{Zn} : 1 < n_{HCl} : 2$ nên Zn dư

$n_{H_2} = n_{Zn} = 0,1(mol)$
$V_{H_2} = 0,1.22,4 = 2,24(l)$

a+b+c) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)=n_{ZnCl_2}=n_{H_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)

d) PTHH: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Theo PTHH: \(n_{H_2}=n_{Cu}=0,1\left(mol\right)\)

\(\Rightarrow m_{Cu}=0,1\cdot64=6,4\left(g\right)\)

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

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

Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow n_{H_2}=0,1.22,4=2,24\left(l\right)\)

c, Ta có: \(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)

d, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

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

\(\Rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)

Bạn tham khảo nhé!