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

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

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)

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

0,1       0,2                        0,1  ( mol )

\(m_{Zn}=0,1.65=6,5g\)
\(C_{M_{HCl}}=\dfrac{0,2}{0,25}=0,8M\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,1        0,2                     0,1 
\(m_{Zn}=0,1.65=6,5g\\ C_{M\left(HCl\right)}=\dfrac{0,2}{0,25}=0,8M\)

n_H2=4.48/22.4=0.2 (mol)

2CH₃COOH + Zn \(\rightarrow\) (CH₃COO)₂Zn + H₂

m_CH3COOH=2.0,2.60=24 (g)

C%= 24/200 .100 %=12%

m_Zn=0,2 . 65=13 (g)

m_(CH3COO)2Zn= 0,2 . 183=36.6 (g)

Chúc bạn học tốt <3

\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{3,36}{22,4}=0,15mol\)

a. PTHH: Fe + H₂SO₄ \(\rightarrow\) FeSO₄ + H₂

       TL:     1         1              1           1

      mol:  0,15  \(\leftarrow\) 0,15 \(\leftarrow\) 0,15  \(\leftarrow\) 0,15

\(b.m_{Fe}=n.M=0,15.56=8,4g\)

Đổi 150ml = 0,15 l

\(c.C_{MddH_2SO_4}=\dfrac{n}{V}=\dfrac{0,15}{0,15}=1M\)

a) $Zn + 2HCl \to ZnCl_2 + H_2$

b) n Zn = n H2 = 4,48/22,4 = 0,2(mol)

m Zn = 0,2.65 = 13(gam)

c) n HCl = 2n H2 = 0,4(mol)

=> C% HCl = 0,4.36,5/200 .100% = 7,3%

Bài 7:

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

b) Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)=n_{Zn\left(pư\right)}\)

\(\Rightarrow m_{Zn\left(pư\right)}=0,2\cdot65=13\left(g\right)\)

c) Theo PTHH: \(n_{HCl}=2n_{H_2}=0,4mol\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)

(Coi như thể tích dd thay đổi không đáng kể)

PTHH: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)

Gộp cả phần a và b

Ta có: \(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)=n_{Ca\left(OH\right)_2}=n_{CaCO_3}\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\\m_{CaCO_3}=0,05\cdot100=5\left(g\right)\end{matrix}\right.\)

Gọi \(n_{Fe}=x\left(mol\right)\)\(;n_{Zn}=y\left(mol\right)\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)

Ta có:  \(\left\{{}\begin{matrix}56x+65y=18,6\\2x+2y=2n_{H_2}=0,6\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\%m_{Fe}=\dfrac{0,1\cdot56}{18,6}\cdot100\%=30,11\%\)

\(\%m_{Zn}=100\%-30,11\%=69,89\%\)

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

0,1     0,2

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

0,2     0,4

\(n_{HCl}=0,2+0,4=0,6mol\)

\(C_M=\dfrac{n}{V}=\dfrac{0,6}{0,2}=3M\)

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

a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\)

\(V_{HCl}=\dfrac{n}{C_M}=\dfrac{0,4}{2}=0,2\left(l\right)\)

b) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,3.65=19,5\left(g\right)\)

c, \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,1}=6\left(M\right)\)