\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\a, Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,V\text{ì}:\dfrac{0,2}{1}>\dfrac{0,1}{1}\Rightarrow Zn\text{dư}\\ \Rightarrow n_{Zn\left(p.\text{ứ}\right)}=n_{ZnCl_2}=n_{H_2}=0,1\left(mol\right)\\b, m_{Zn\left(p.\text{ứ}\right)}=0,1.65=6,5\left(g\right)\\ n_{HCl}=0,1.2=0,2\left(mol\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ d,m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)

a) Zn + 2HCl --> ZnCl₂ + H₂

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

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

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

=> V_H2 = 0,2.22,4 = 4,48 (l)

c) \(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,2}{1}\) => CuO dư, H₂ hết

PTHH: CuO + H₂ --t^o--> Cu + H₂O

             0,2<--0,2-------->0,2

=> n_CuO(dư) = 0,3 - 0,2 = 0,1 (mol)

m_Cu = 0,2.64 = 12,8 (g)

\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.n_{Zn}=\dfrac{65}{65}=1\left(mol\right)\\ n_{ZnCl_2}=n_{Zn}=1\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=1.136=136\left(g\right)\\ c.n_{H_2}=n_{Zn}=1\left(mol\right)\\ \Rightarrow V_{H_2}=1.22,4=22,4\left(l\right)\)

a) Zn + 2HCl --> ZnCl₂ + H₂

b) Theo ĐLBTKL: m_Zn + m_HCl = m_ZnCl2 + m_H2

=> m_H2 = 5,2 + 5,84 - 10,88 = 0,16 (g)

c) \(n_{H_2}=\dfrac{0,16}{2}=0,08\left(mol\right)\)

=> V_H2 = 0,08.22,4 = 1,792(l)

a. \(n_{Zn}=\dfrac{6.5}{65}=0,1\left(mol\right)\)

PTHH : Zn + 2HCl -> ZnCl₂ + H₂

            0,1      0,2                  0,1

b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

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

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

0,1      0,2                       0,1

\(V_{H_2}=0,1\cdot22,4=2,24l\)

\(m_{HCl}=0,2\cdot36,5=7,3g\)

a) PTHH : Zn + 2 HCl -> ZnCl2 + H2 

nH2= 0,15(mol)

-> nZn= nZnCl2=nH2= 0,15(mol)

-> Số nguyên tử kẽm p.ứ: 6.10²³ .0,15= 9.10²² (nguyên tử) 

b) mZnCl2= 136.0,15= 20,4(g)

c) H2+ 1/2 O2 -to-> H2O 

nH2= 0,15/2= 0,075(mol)

-> nH2O= nH2= 0,075(mol)

-> mH2O= 0,075.18= 1,35(g)

a)

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

b),c)

Theo PTHH :

\(n_{ZnCl_2} = n_{H_2} = n_{Zn} = \dfrac{13}{65} = 0,2(mol)\)

Vậy : 

\(m_{ZnCl_2} = 0,2.136 = 27,2(gam)\\ V_{H_2} =0,2.22,4 = 4,48(lít)\)

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

b+c)

Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)=n_{ZnCl_2}=n_{H_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,2\cdot136=27,2\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\end{matrix}\right.\)

a. Zn + 2HCl → ZnCl₂ + H₂

b. n_Zn = n\(_{ZnCl_2}\) =\(\dfrac{13}{65}=0,2\left(mol\right)\) => m\(_{ZnCl_2}\)= 0,2.136 = 27,2(g)

c. n\(_{H_2}\)= n_Zn = 0,2 (mol) => V\(_{H_2}\)=0,2.22,4 = 4,48 (lít)

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

b) \(n_{ZnCl_2}=\dfrac{20,4}{136}=0,15\left(mol\right)\)

\(n_{H_2}=n_{ZnCl_2}=0,15\left(mol\right)\)

\(V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)

c) \(m_{H_2}=0,15.2=0,3\left(g\right)\)

\(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)

\(m_{HCl}=0,3.36,5=10,95\left(g\right)\)

\(m_{Zn}=20,4+0,3-10,95=9,75\left(g\right)\)

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

ap dung DLBTKL ta co

\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\)

\(=>m_{H_2}=m_{Zn}+m_{HCl}-m_{ZnCl_2}\\ =>m_{H_2}=13+14,6-27,2\\ =>m_{H_2}=0,4\left(g\right)\)