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

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!

\(^nZnCl_2=\dfrac{20,4}{136}=0,15\left(mol\right)\)

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

mol         0,15                   0,15        0,15

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

c)   \(m=^mZn=0,15.65=9,75\left(g\right)\)

Chúc bạn học tốt!!! 

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 + 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 --> 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) PTHH:

Zn + 2HCl  ZnCl₂ + H₂
 

a, PTHH: Zn + 2HCl \(\rightarrow\) ZnCl₂ + H₂

b, \(m_{Zn}=0,1.65=6,5g\)

\(m_{H_2}=\left(\dfrac{2,24}{22,4}\right).2=0,2g\)

Theo ĐLBTKL, ta có:

m_Zn + m_HCl = m_{\(ZnCl_2\)} + m_{\(H_2\)}

\(m_{ZnCl_2}=\left(6,5+72\right)-0,2=78,3g\)

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

\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

\(\Leftrightarrow n_{ZnCl_2}=0.2\left(mol\right)\)

\(m_{ZnCl_2}=0.2\cdot136=27.2\left(g\right)\)

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

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

0,1                     0,1         0,1 

\(b,C_M=\dfrac{0,1}{0,1}=1M\)

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