\(n_{HCl}=0,25.2=0,5\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{Zn}=0,25.65=16,25\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)

Không biết đúng không nữa;-;;;

a) PTHH: Zn + 2HCl -> ZnCl2 + H2 

b) HCl=250ml=0,25l

n2HCl= V/22,4= 0,5/22,4= 0,02(mol)

Zn  +  2HCl -> ZnCl2  +  H2

1          2              1         1

0,01 <-0,5--------------> 0,01

mZn= n.M= 0,01.65= 0,65(gam)

c) VH2=n . 22,4= 0,01 . 22,4= 0,224(l)
 

a, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Zn + 2HCl -----> ZnCl₂ + H₂

0,2     0,4               0,2     0,2

b, \(m_{Zn}=0,2.65=13\left(g\right)\)

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

c, \(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

\(a,n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

         0,2<--0,4<--------0,2<---0,2

\(b,\left\{{}\begin{matrix}m_{Zn}=0,1.65=13\left(g\right)\\m_{HCl}=0,4.36,5=14,6\left(g\right)\end{matrix}\right.\\ c,m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

a) Zn + HCl -> ZnCl ₂ + H₂

Zn + 2HCl -> ZnCl₂ + H₂

1 : 2 : 1 : 1
c) m_Zn + M_HCl = m_ZnCl + m_H

130 + m_HCl =2,72 +4

m_HCl =(2,72+4) - 130

Còn lại bạn tự tính nhé !

a) Zn + HCl ----> ZnCl2 + H2

Zn + 2HCl -> ZnCl2 + H2

b) Zn + 2HCl -> ZnCl2 +H2

1ntử : 2 ptử : 1 ptử : 1 ptử

c) Áp dụng ĐLBTKL, ta có:

m_Zn + m_HCl = m_ZnCl2 + m_H2

\(\Rightarrow\)m_HCl= (m_ZnCl2 + m_H2) - m_Zn

_{\(\Rightarrow\)}m_{HCl= (2,72 + 4) -1,3}

\(\Rightarrow\)m_{HCl= 5,42}

Hình như m_Zn bạn bị nhầm nên mik lm thành 1,3g

a) Zn + 2HCl →ZnCl₂  + H₂

b) nZn = 6,5/65 = 0,1 mol . Theo tỉ lệ pư => nH₂ = nZn = nZnCl₂ =0,1 mol <=> V_H2(đktc) = 0,1.22,4 = 2,24 lít.

c) mZnCl₂ = 0,1 . 136 = 13,6 gam

d) nHCl =2nZn = 0,2 mol => mHCl = 0,2.36,5= 7,3 gam

Cách 2: áp dụng định luật BTKL => mHCl = mZnCl₂ + mH₂ - mZn 

<=> mHCl = 13,6 + 0,1.2 - 6,5 = 7,3 gam

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

b. Theo ĐLBTKL, ta có:

\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ b.\Leftrightarrow6,5+7,3=13,6+m_{H_2}\\ \Leftrightarrow m_{H_2}=\left(6,5+7,3\right)-13,6=0,2\left(g\right)\)

Chúc em học tốt!

a) Zn + 2HCl --> ZnCl₂ + H₂

Phản ứng thế

b) \(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

_____0,02------------>0,02-->0,02

=> m_ZnCl2 = 0,02.136 = 2,72(g)

=> V_H2 = 0,02.22,4 = 0,448(l)

a, Bảo toàn e:

\(2n_{Zn}=2n_{H_2}\)

\(\Leftrightarrow n_{Zn}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{Zn}=19,5\left(g\right)\)

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

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

Theo gt ta có: $n_{Zn}=0,1(mol)$

a, $Zn+2HCl\rightarrow ZnCl_2+H_2$

b, Ta có: $n_{H_2}=0,1(mol)\Rightarrow V_{H_2}=2.24(l)$

c, Ta có: $n_{HCl}=2.n_{Zn}=0,2(mol)\Rightarrow m_{HCl}=7,3(g)$

\(a)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ b)\\ n_{H_2} = n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)\\ \Rightarrow V_{H_2} = 0,1.22,4 = 2,24(lít)\\ c)\\ n_{HCl} = 2n_{Zn} = 0,1.2 = 0,2(mol)\\ \Rightarrow m_{HCl} = 0,2.36,5 = 7,3\ gam\)