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

\(n_{Zn}=\dfrac{26}{65}=0.4\left(mol\right)\)

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

\(0.4.......0.8....................0.4\)

\(m_{HCl}=0.8\cdot36.5=29.2\left(g\right)\)

\(V_{H_2}=0.4\cdot22.4=8.96\left(l\right)\)

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, Kẽm + axit clohidric → Khí hidro + muối kẽm

N_zn=0,1mol

a) có pt : Zn + 2Hcl -> ZnCl₂ + H₂

1 -> 2 -> 1 -> 1 mol

0,1-> 0,2 -> 0,1 -> o,1 mol

b) số Hcl đã dùng khi pứ là :

0,2 . 36,5= 7,3

-> số gam hcl dư là 10,95- 7,3=3,65(g)

c)m_ZnCl2=0,1 . 136=13,6g

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_{\(H_2\)}

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

$a\big)$

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

$Zn+2HCl\to ZnCl_2+H_2$

Theo PT: $n_{H_2}=n_{Zn}=0,3(mol)$

$\to V_{H_2}=0,3.22,4=6,72(l)$

$b\big)$

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

Theo PT: $n_{Fe}=\dfrac{2}{3}n_{H_2}=0,2(mol)$

$\to m_{Fe}=0,2.56=11,2(g)$

\(a.Mg+HCl\rightarrow MgCl_2+H_2\\ b.n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\\ \Rightarrow m_{MgCl_2}=0,1.95=9,5\left(g\right)\\ c.n_{H_2}=n_{Mg}=0,1\left(mol\right)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

a) Lập công thức hóa học của muối kẽm clorua:  Z n C l 2

b) Khối lượng muối Z n C l 2  = 6,5 + 7,3 – 0,2 = 13,6 (gam)