\(a.PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\)

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

Từ PTHH trên ta có:

1 mol Mg tác dụng với 2 mol HCl sinh ra 1 mol MgCl₂ và 1 mol H₂ 

=> 0,1 mol Mg tác dụng với 0,2 mol HCl sinh ra 0,1 mol MgCl₂ và 0,1 mol H₂ 

\(\Rightarrow m_{HCl}=36,5.0,2=7,3\left(g\right)\)

\(c.\Rightarrow m_{H_2}=0,1.2=0,2\left(g\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)

\(d.C_1:\\ Áp.dụng.ĐLBTKL,ta.có:\\ m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\\ \Rightarrow m_{MgCl_2}=m_{Mg}+m_{HCl}-m_{H_2}=2,4+7,3-0,2=9,5\left(g\right)\)

\(C_2:Từ.PTHH.trên.ta.có:\\ n_{MgCl_2}=0,1\left(mol\right)\\ \Rightarrow m_{MgCl_2}=0,1.95=9,5\left(g\right)\)

 PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

Làm gộp các phần còn lại

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1mol\\n_{H_2SO_4}=n_{H_2}=0,3mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\m_{H_2SO_4}=0,3\cdot98=29,4\left(g\right)\end{matrix}\right.\) 

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

           0,5        0,25       0,25

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(a,m_{ZnCl_2}=0,25.136=34\left(g\right)\)

\(b,m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(m_{ddHCl}=\dfrac{18,25.100}{7,3}=250\left(g\right)\)

\(c,2K+2H_2O\rightarrow2KOH+H_2\uparrow\)

   0,5                                   0,25

\(m_K=39.0,5=19,5\left(g\right)\)

a) PTHH : Zn + 2 HCl -> ZnCl2 + H2 

nH2= 0,15(mol)

-> nZn= nZnCl2=nH2= 0,15(mol)

-> Số nguyên tử kẽm p.ứ: 6.10²³ .0,15= 9.10²² (nguyên tử) 

b) mZnCl2= 136.0,15= 20,4(g)

c) H2+ 1/2 O2 -to-> H2O 

nH2= 0,15/2= 0,075(mol)

-> nH2O= nH2= 0,075(mol)

-> mH2O= 0,075.18= 1,35(g)

Ta có: \(n_{HCl}=\dfrac{200}{1000}.2=0,4\left(mol\right)\)

\(PTHH:Mg+2HCl--->MgCl_2+H_2\uparrow\left(1\right)\)

a. Theo PT₍₁₎: \(n_{Mg}=n_{H_2}=n_{MgCl_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,2.24=4,8\left(g\right)\\V_{H_2}=0,2.22,4=4,48\left(lít\right)\end{matrix}\right.\)

b. \(PTHH:2NaOH+MgCl_2--->Mg\left(OH\right)_2\downarrow+2NaCl\left(2\right)\)

Ta có: \(n_{NaOH}=\dfrac{\dfrac{20\%.100}{100\%}}{40}=0,5\left(mol\right)\)

Ta thấy: \(\dfrac{0,5}{2}>\dfrac{0,2}{1}\)

Vậy NaOH dư.

Theo PT₍₂₎: \(n_{Mg\left(OH\right)_2}=n_{MgCl_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Mg\left(OH\right)_2}=0,2.58=11,6\left(g\right)\)

a: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

200ml=0,2 lít

\(n_{HCl}=0.2\cdot22.4=4.48\left(mol\right)\)

\(\Leftrightarrow n_{H_2}=2.24\left(mol\right)\)

\(\Leftrightarrow m_{H_2}=n_{H_2}\cdot M=2.24\cdot1=2.24\left(g\right)\)

\(n_{MgCl_2}=2.24\left(mol\right)\)

\(\Leftrightarrow n_{Mg}=2.24\left(mol\right)\)

\(\Leftrightarrow m_{Mg}=2.24\cdot24=53.76\left(g\right)\)

a, \(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)

PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

b, Theo PT: \(n_{FeSO_4}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{FeSO_4}=0,15.152=22,8\left(g\right)\)

c, \(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,15}{1}\), ta được CuO dư.

Theo PT: \(n_{Cu}=n_{CuO\left(pư\right)}=n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow n_{CuO\left(dư\right)}=0,3-0,15=0,15\left(mol\right)\)

Chất rắn thu được sau pư gồm Cu và CuO dư.

⇒ m _{chất rắn} = m_Cu + m_{CuO (dư)} = 0,15.64 + 0,15.80 = 21,6 (g)

\(n_{Zn}=\dfrac{9.75}{65}=0.15\left(mol\right)\)

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

\(0.15.....0.3...................0.15\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)

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

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

______0,2_________________0,2 (mol)

b, V_H2 = 0,2.22,4 = 4,48 (l)

c, Ta có: \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)

PT: \(FeO+H_2\underrightarrow{t^o}Fe+H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\), ta được H₂ dư.

Theo PT: \(n_{Fe}=n_{FeO}=0,1\left(mol\right)\)

⇒ m_Fe = 0,1.56 = 5,6 (g)

Bạn tham khảo nhé!

a) Zn  + 2HCl \(\rightarrow\)ZnCl₂ + H₂

b) m_Zn = \(\dfrac{13}{65}\)=0,2 (mol)

         Zn  + 2HCl \(\rightarrow\)ZnCl₂ + H₂

(mol) 0,2 ----------------------> 0,2

\(V_{H_2}\)= 0,2 . 22,4 = 4,48(lít)

c)\(n_{FeO}\)=\(\dfrac{7,2}{72}\)=0,1 (mol)

         H₂ + FeO \(\underrightarrow{t^o}\)Fe + H₂O

(mol)         0,1----->0,1

m_Fe = 0,1 . 56 = 5,6(g)