a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

nZn = \(\dfrac{13}{65}=0,2mol\)

Theo pt: nH₂ = nZn = 0,2 mol

=> VH₂ = 0,2.22,4 = 4,48 lít

b) Theo pt: nZnCl₂ = nZn = 0,2 mol

=> mZnCl₂ = 0,2.136 = 27,2g

a)

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

b)

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

Ta thấy : \(\dfrac{n_{Zn}}{1} = 0,2 > \dfrac{n_{HCl}}{2} = 0,15\) nên Zn dư.

Theo PTHH :

\(n_{Zn\ pư} = 0,5n_{HCl} = 0,15(mol)\\ \Rightarrow n_{Zn\ dư} = 0,2 - 0,15 = 0,05(mol)\\ \Rightarrow m_{Zn\ dư} = 0,05.65 = 3,25(gam)\)

c)

Ta có :

\(n_{H_2} = n_{Zn\ pư} = 0,15(mol)\\ \Rightarrow V_{H_2} = 0,15.22,4 = 3,36(lít)\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{18,25}{35}=0,5\left(mol\right)\\ a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,5}{2}>\dfrac{0,2}{1}\Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,5-0,2.2=0,1\left(mol\right)\\ m_{HCl}=0,1.36,5=3,65\left(g\right)\\ c.n_{ZnCl_2}=n_{Zn}=n_{H_2}=0,2\left(mol\right)\\ m_{ZnCl_2}=0,2.136=27,2\left(g\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)

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

\(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

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

            1          2            1            1

           0,2       0,5         0,2         0,2

b) Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\)

                 ⇒ Zn phản ứng hết , HCl dư

                 ⇒ Tính toán dựa vào số mol của Zn

\(n_{HCl\left(dư\right)}=0,5-\left(0,2.2\right)=0,1\left(mol\right)\)

⇒ \(m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)

c) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)

\(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

 Chúc bạn học tốt

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

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

\(0.2....................0.2..........0.2\)

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

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

Ulatr thanks mn nhìu ♥️

\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=2\cdot0,15=0,3\left(mol\right)\)

Vì \(\dfrac{n_{Zn}}{1}>\dfrac{n_{HCl}}{2}\) nên sau p/ứ Zn dư

\(\Rightarrow n_{Zn}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow m_{Zn}=0,15\cdot65=9,75\\ \Rightarrow m_{Zn\left(dư\right)}=13-9,75=3,25\left(g\right)\\ c,n_{H_2}=n_{Zn}=0,15\left(mol\right)\\ \Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)

\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.n_{Zn}=n_{ZnCl_2}=0,1\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=0,1.136=13,6\left(g\right)\\ c.n_{H_2}=n_{Zn}=0,1\left(mol\right)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

a) Zn + 2HCl → ZnCl₂  + H₂ 
b) mZnCl₂ = 0,1 . 136 = 13,6 gam
c) 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.

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

Phương trình hóa học

Zn + 2HCl --> ZnCl₂ + H₂ 

1   :   2       :       1     :  1 

0,1                      0,1    0,1 

mol                     mol    mol

\(V_{H_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)

\(m_{ZnCl_2}=n.M=0,1.136=13,6\left(g\right)\)

\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)

\(n_{HCl}=\dfrac{100\cdot14.6\%}{36.5}=0.4\left(mol\right)\)

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

\(1........2\)

\(0.1......0.4\)

\(LTL:\dfrac{0.1}{1}< \dfrac{0.4}{2}\Rightarrow HCldư\)

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{\text{dung dịch sau phản ứng}}=6.5+100-0.1\cdot2=106.3\left(g\right)\)

\(C\%ZnCl_2=\dfrac{0.1\cdot136}{106.3}\cdot100\%=12.79\%\)

\(C\%HCl\left(dư\right)=\dfrac{\left(0.4-0.2\right)\cdot36.5}{106.3}\cdot100\%=6.87\%\%\)

a, \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

\(m_{HCl}=36,5.15\%=5,475\left(g\right)\Rightarrow n_{HCl}=\dfrac{5,475}{36,5}=0,15\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,15}{2}\), ta được Mg dư.

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,075\left(mol\right)\Rightarrow V_{H_2}=0,075.22,4=1,68\left(l\right)\)

b, \(n_{Mg\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,075\left(mol\right)\Rightarrow n_{Mg\left(dư\right)}=0,1-0,075=0,025\left(mol\right)\)

\(\Rightarrow m_{Mg\left(dư\right)}=0,025.24=0,6\left(g\right)\)

c, - Cách 1:

\(n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,075\left(mol\right)\Rightarrow m_{MgCl_2}=0,075.95=7,125\left(g\right)\)

- Cách 2: 

Theo ĐLBT KL, có: m_{Mg (pư)} + m_HCl = m_MgCl2 + m_H2

⇒ m_MgCl2 = 2,4 - 0,6 + 5,475 - 0,075.2 = 7,125 (g)

8====D

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