Câu 1
\(a)PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b)200ml=0,2l\\ n_{HCl}=0,2.1=0,2mol\\ n_{H_2}=n_{MgCl_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}\cdot0,2=0,1mol\\ V_{H_2}=0,1.24,79=2,479l\\ c)C_{M_{MgCl_2}}=\dfrac{0,1}{0,2}=0,5M\)

Câu 2
\(a)PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b)n_{Mg}=\dfrac{1,2}{24}=0,05mol\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,05mol\\ V_{H_2}=0,05.24,79=1,2395l\\ c)n_{HCl}=2n_{Mg}=2.0,05=0,1mol\\ V_{ddHCl}=\dfrac{0,1}{2}=0,05l\\ d)C_{M_{MgCl_2}}=\dfrac{0,05}{0,05}=1M\)

là giúp minhd có ô-xy để thở

Không khí có vai trò quan trọng trong đời sống, giúp vạn vật phát triển, duy trì mọi hoạt động sống của tế bào. Tóm lại không khí duy trì sự sống của mọi loại sinh vật. Hết

\(n_{BaCl_2}=\dfrac{200.20,8\%}{208}=0,2\left(mol\right)\\ PTHH:BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ n_{BaSO_4}=n_{H_2SO_4}=n_{BaCl_2}=0,2\left(mol\right)\\ a,m_{kt}=m_{BaSO_4}=233.0,2=46,6\left(g\right)\\ b,C\%_{ddH_2SO_4}=\dfrac{0,2.98}{200}.100\%=9,8\%\)

\(n_{SO_2}=\dfrac{11,2}{22,4}=0,5mol\\ n_{Ca\left(OH\right)_2}=0,2.0,5=0,1mol\\ SO_2+Ca\left(OH\right)_2\rightarrow CaSO_3+H_2O\\ \Rightarrow\dfrac{0,5}{1}>\dfrac{0,1}{1}\Rightarrow SO_2.dư\\ n_{SO_2.pứ}=n_{Ca\left(OH\right)_2}=0,1mol\\ m_{SO_2.dư}=\left(0,5-0,1\right).64=25,6g\)

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

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Theo pt: \(n_{HCl}=2n_{Fe}=0,2\left(mol\right)\)

\(\Rightarrow C_MHCl=\dfrac{0,2}{0,5}=0,4M\)

Fe + 2HCl → FeCl₂ + H₂

n_Fe = \(\dfrac{5,6}{56}\) = 0,1 (mol)

Theo PT:n_HCl = 2n_Fe = 0,2 (mol)

⇒ C_MHCl = \(\dfrac{0,2}{0,5}\) = 0,4M

\(Fe\left(NO_3\right)_3+3KOH\rightarrow Fe\left(OH\right)_3+3KNO_3\\ 2Fe\left(OH\right)_3\rightarrow\left(t^o\right)Fe_2O_3+3H_2O\\ Fe_2O_3+3CO\rightarrow\left(t^o\right)2Fe+3CO_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ FeCl_2+2KOH\rightarrow Fe\left(OH\right)_2+2KCl\)

Sửa đề chất cuối thành Fe(OH)₂

\(Muối:ACO_3,B_2CO_3\\ ACO_3+2HCl\rightarrow ACl_2+CO_2+H_2O\\ B_2CO_3+2HCl\rightarrow2BCl+CO_2+H_2O\\ n_{CO_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\Rightarrow m_{CO_2}=44.0,03=1,32\left(g\right)\\ n_{CO_3^{2-}}=n_{CO_2}=0,03\left(mol\right);n_{Cl^-}=2.n_{CO_2}=0,03.2=0,06\left(mol\right)\\ m_{muối.ddX}=m_{muối.cacbonat}+m_{Cl^-}-m_{CO_3^{2-}}=10+0,06.35,5-60.0,03=10,3\left(g\right)\)

\(a,Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{ZnCl_2}=n_{H_2}=n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ b,m_{ZnCl_2}=136.0,1=13,6\left(g\right);V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c,Vì:\dfrac{0,1}{1}>\dfrac{0,2}{1}\Rightarrow n_{Zn\left(TT\right)}=0,1\left(mol\right);n_{Zn\left(LT\right)}=n_{ZnCl_2}=0,2\left(mol\right)\\ H=\dfrac{0,1}{0,2}.100\%=50\%\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{ZnCl_2}=\dfrac{20,6}{136}\approx0,15147\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,2}{1}>\dfrac{0,15147}{1}\\ Có:m_{ZnCl_2\left(LT\right)}=0,2.136=27,2\left(g\right)\\ H=\dfrac{20,6}{27,2}.100\%\approx75,735\%\)

a)

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

b)

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

Theo PT:\(n_{H_2}=n_{Zn}=0,2mol\)

\(\Rightarrow V_{H_2}=0,2.24,79=4,958l\)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)