nHCl = 0.3*0.75 = 0.225 (mol) 

Zn + 2HCl => ZnCl2  + H2 

0.1125..0.225....0.1125..0.1125

mZn = 0.1125*65 = 7.3125 (g) 

VH2 = 0.1125*22.4 = 2.52 (l) 

CM ZnCl2 = 0.1125/0.3 = 0.375 (M)

nHCl= 0,75 x 0,3= 0,225(mol)

PTHH: Zn +  2HCl -> ZnCl2 + H2

a) 0,1125____0,225___0,1125___0,1125(mol)

=> a=mZn=0,1125.65=7,3125(g)

V(H2,đktc)=0,1125 x 22,4=2,52(l)

b) VddZnCl2=VddHCl=0,3(l)

=>CMddZnCl2=0,1125/0,3=0,375(M)

a. \(n_{SO_3}=\dfrac{8}{80}=0,1\left(mol\right)\)

\(C_M=\dfrac{0,1}{200}=0,0005\left(mol\text{/}l\right)\) 

a, PTPƯ: SO₃ + H₂O ---> H₂SO₄

n_SO3=\(\dfrac{2,24}{22,4}=0,1mol\)

1 mol SO₃ ---> 0,1 mol H₂SO₄

nên 0,1 mol SO₃ ---> 0,1 mol H₂SO₄

C_{M H2SO4}=\(\dfrac{0,1}{0,5}\)=0,2 M

b, PTPƯ: Zn + H₂SO₄ ---> ZnSO₄ + H₂

1 mol H₂SO₄ ---> 1 mol Zn

nên 0,1 mol H₂SO₄ ---> 0,1 mol Zn

m_Zn=0,1.65=6,5 g

H₂+Cl₂->2HCl

         0,35---0,7 mol

n _H2=\(\dfrac{8,96}{22,4}\)=0,4 mol

n _Cl2=\(\dfrac{7,84}{22,4}\)=0,35 mol

=>H2 dư :0,05 mol

=>V_HCl=0,7.22,4=15,68l

b) C% _HCl=\(\dfrac{0,7.36,5}{25,55+224,45}\).100=10,22%

a: \(H_2+Cl_2\rightarrow2HCl\)

\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

\(n_{Cl_2}=\dfrac{7.84}{22.4}=0.35\left(mol\right)\)

=>Cl2 thiếu, H2 dư

\(V_{HCl}=2\cdot V_{Cl_2}=0.7\left(mol\right)\)

a)

$2K + 2H_2O \to 2KOH + H_2$

b)

n K = 3,9/39 = 0,1(mol)

Theo PTHH :

n H2 = 1/2 n K = 0,05(mol)

=> V H2 = 0,05.22,4 = 1,12(lít)

c)

m dd sau pư = m K + m nước - m H2 = 3,9 + 36,2 - 0,05.2 = 40(gam)

C% KOH = 0,1.56/40 .100% =14%

Câu 5. a) \(SO_2+\dfrac{1}{2}O_2-^{t^o,V_2O_5}\rightarrow SO_3\)

\(n_{SO_3}=n_{SO_2}=0,1\left(mol\right)\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(n_{H_2SO_4}=n_{SO_3}=0,1\left(mol\right)\)

=> \(CM_{H_2SO_4}=\dfrac{0,1}{0,2}=0,5M\)

b) \(n_{Zn}=0,05\left(mol\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

Lập tỉ lệ : \(\dfrac{0,05}{1}< \dfrac{0,1}{1}\)=> Sau phản ứng H2SO4 dư

=> \(m_{H_2SO_4\left(dư\right)}=\left(0,1-0,05\right).98=4,9\left(g\right)\)

Câu 5 . \(n_{Al_2O_3}=0,2\left(mol\right);n_{H_2SO_4}=0,8\left(mol\right)\)

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

Lập tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,8}{3}\) => Sau phản ứng H2SO4 dư

\(m_{H_2SO_4}=\left(0,8-0,2.3\right).98=19,6\left(g\right)\)

b)\(n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,2\left(mol\right)\)

\(m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\)

Câu 6. a) \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(n_{Al}=0,05\left(mol\right);n_{H_2SO_4}=\dfrac{200.3,92\%}{98}=0,08\left(mol\right)\)

Lập tỉ lệ : \(\dfrac{0,05}{2}< \dfrac{0,08}{3}\)=> Sau phản ứng H2SO4 dư

b) \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,075\left(mol\right)\)

=> \(V_{H_2}=0,075.22,4=1,68\left(l\right)\)

c) \(m_{ddsaupu}=1,35+200-0,075.2=201,2\left(g\right)\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(0,08-0,075\right).98}{201,2}.100=0,243\%\)

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,025.342}{201,2}.100=4,25\%\%\)

a) \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

PTHH: 2Na + 2H₂O ---> 2NaOH + H₂ => ddA là NaOH

            0,2----------------->0,2------>0,1

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

c) \(C_{M\left(NaOH\right)}=\dfrac{0,2}{0,4}=0,5M\)

a) 

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

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,2-->0,4----->0,2--->0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

b) m_HCl = 0,4.36,5 = 14,6 (g)

=> \(m_{dd.HCl}=\dfrac{14,6.100}{7,3}=200\left(g\right)\)

c)

m_{dd sau pư} = 13 + 200 - 0,2.2 = 212,6 (g)

m_ZnCl2 = 0,2.136 = 27,2 (g)

=> \(C\%=\dfrac{27,2}{212,6}.100\%=12,8\%\)

a) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

           0,4-->0,6---------->0,2------->0,6

=> \(C_{M\left(dd.H_2SO_4\right)}=\dfrac{0,6}{0,15}=4M\)

b) V_H2 = 0,6.22,4 = 13,44 (l)

c) \(C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M\)

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) 
           0,4     0,6                   0,2             0,6 
\(C_M_{H_2SO_4}=\dfrac{0,6}{0,15}=4M\\ V_{H_2}=0,622,4=13,44L\) 
\(C_M=\dfrac{0,2}{0,15}=1,3M\)

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ m_{HCl}=\dfrac{109,5\cdot10\%}{100\%}=10,95\left(g\right)\\ \Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\\ a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ \text{Vì }\dfrac{n_{Mg}}{1}< \dfrac{n_{HCl}}{2}\text{ nên sau p/ứ }HCl\text{ dư}\\ \Rightarrow n_{H_2}=0,1\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,1\cdot22,4=2,24\left(l\right)\)

\(b,n_{MgCl_2}=0,1\left(mol\right)\\ \Rightarrow m_{CT_{MgCl_2}}=0,1\cdot95=9,5\left(g\right)\\ m_{H_2}=0,1\cdot2=0,2\left(mol\right)\\ m_{dd_{MgCl_2}}=2,4+109,5-0,2=111,7\left(g\right)\\ \Rightarrow C\%_{MgCl_2}=\dfrac{9,5}{111,7}\cdot100\%\approx8,5\%\)