nBa=0,3 mol

mH2SO4=9,8g=>nh2SO4=0,1mol

PTHH: Ba+H2SO4=>BaSO4+H2

           0,3mol:0,1mol

=> nBa dư theo n H2SO4

p/ư:    0,1mol<-0,1mol->0,1mol->0,1mol

thể tích H2: V=0,1.22,4=2,24ml

theo đl btoan khối lượng ta có : mBaSO4= mBa+mH2SO4-mH2=41,1+200-0,1.2=240,9g

mBaSO4= 0,1.233=23,3g

=> C% BaSO4=9,67%

a) Ba + H2SO4 ---> BaSO4+ H2

nBa=41,1/137=0,3 mol , mH2SO4= [200 * 4,9]/100=9,8 gam => nH2SO4=0,1 mol

=> theo pt và bài cho H2SO4 phản ứng hết và Ba dư

nH2=nH2SO4=0,1=> VH2=0,1*22,4=2,24 lít

b) mdd sau phản ứng =mBa+ mddH2SO4-mH2-mBa dư=213,5 gam

mBaSO4=23,3 gam

=> C% BaSO4= [23,3/213,5]*100%=10,91%.

a)Đổi \(V_{H_2SO_4}=100ml=0,1l\)

Số mol của 2,7 gam Al:

\(n_{Al}=\dfrac{m}{M}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

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

Tỉ lệ     2      :  3          :      1           :         3

          0,1 ->  0,15       :      0,05     :         0,15(mol)

Nồng độ mol của dung dịch H2SO4:

\(C_{M_{H_2SO_4}}=\dfrac{n_{H_2SO_4}}{V_{H_2SO_4}}=\dfrac{0,15}{0,1}=1,5\left(M\right)\)

b) thể tích của 0,15 mol H2:

\(V_{H_2}=n.22,4=0,15.22,4=3,36\left(l\right)\)

c) nồng độ mol của dd \(Al_2\left(SO_4\right)_3\) :

\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{n}{V}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)

Bài 3 : 

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

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

        2            3                   1               3

       0,1       0,15                 0,05       0,15

a) \(n_{H2}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)

\(m_{H2}=0,15.2=0,3\left(g\right)\)

\(V_{H2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)

b) \(n_{H2SO4}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)

⇒ \(m=0,15.98=14,7\left(g\right)\)

\(C_{ddH2SO4}=\dfrac{14,7.100}{200}=7,35\)⁰/₀

c) \(n_{Al2\left(SO4\right)3}=\dfrac{0,15.1}{3}=0,05\left(mol\right)\)

⇒ \(m_{Al2\left(SO4\right)3}=0,05.342=17,1\left(g\right)\)

\(m_{ddspu}=2,7+200-0,3=302,4\left(g\right)\)

\(C_{Al2\left(SO4\right)3}=\dfrac{17,1.100}{302,4}=5,65\)⁰/₀

 Chúc bạn học tốt

a)

$Mg + H_2SO_4 \to MgSO_4 + H_2$

Theo PTHH : 

$n_{H_2} = n_{Mg} = \dfrac{9,6}{24} = 0,4(mol)$
$V_{H_2} = 0,4.22,4 = 8,96(lít)$

b)

$n_{H_2SO_4} = n_{Mg} =0,4(mol) \Rightarrow a = \dfrac{0,4.98}{9,8\%} = 400(gam)$
Sau phản ứng : 

$m_{dd} = 9,6 + 400 - 0,4.2 = 408,8(gam)$
$C\%_{MgSO_4} = \dfrac{0,4.120}{408,8}.100\% = 11,74\%$

a,n_Mg=9,6/24=0,4(mol)

PTHH: Mg + H₂SO₄ → MgSO₄ + H₂

Mol:     0,4       0,4            0,4       0,4

\(\Rightarrow V_{H_2}=0,4.22,4=8,96\left(l\right)\)

b,\(m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=39,2:9,8\%=400\left(g\right)\)

\(C\%_{MgSO_4}=\dfrac{0,4.120}{400+9,6}.100\%=11,79\%\)

\(Mg+H_2SO_4 \to MgSO_4+H_2\\ n_{Mg}=\frac{9,6}{24}=0,4mol\\ n_{Mg}=n_{H_2}=n_{H_2SO_4}=n_{MgSO_4}=0,4mol\\ a/\\ V_{H_2}=0,4.22,4=8,96(l)\\ b/\\ m_{H_SO_4(dd)}=\frac{0,4.98}{9,8\%}=400g\\ C\%=\frac{0,4.120}{408,8}.100=11,74\%\)

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ Mg+H_2SO_4\to MgSO_4+H_2\\ MgO+H_2SO_4\to MgSO_4+H_2O\\ \Rightarrow n_{Mg}=0,25(mol)\\ a,\begin{cases} \%_{Mg}=\dfrac{0,25.24}{14}.100\%=42,86\%\\ \%_{MgO}=100\%-42,86\%=57,14\% \end{cases}\\ b,n_{MgO}=\dfrac{14-0,25.24}{40}=0,2(mol)\\ \Rightarrow \Sigma n_{H_2SO_4}=0,2+0,25=0,45(mol)\\ \Rightarrow C\%_{H_2SO_4}=\dfrac{0,45.98}{200}.100\%=22,05\%\)

a,\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Mol:     0,1        0,15           0,05           0,15

\(m_{H_2}=0,15.2=0,3\left(g\right)\)

\(V_{H_2}=0,15.24,79=3,7185\left(l\right)\)

b, \(m_{ddH_2SO_4}=\dfrac{0,15.98.100\%}{200}=7,35\%\)

c, m_{dd sau pứ} = 2,7 + 200 - 0,3 = 202,4 (g)

\(C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342.100\%}{202,4}=8,45\%\)

ai đi ngang giúp mk vs :((

ai đi ngang giúp a

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\\ pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\) 
  \(LTL:\dfrac{0,2}{1}< \dfrac{0,25}{1}\) 
=> H₂SO₄ dư 
\(n_{H_2}=n_{H_2SO_4\left(p\text{ư}\right)}=n_{Fe}=0,2\left(mol\right)\\ V_{H_2}=0,2.22,4=4,48l\\ m_{H_2SO_4\left(d\right)}=\left(0,25-0,2\right).98=4,9g\) 
 

c

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ a,V_{H_2\left(Đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b,m_{ZnCl_2}=136.0,2=27,2\left(g\right)\\ c,n_{HCl}=0,2.2=0,4\left(mol\right)\\ C\%_{ddHCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)

\(m_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ m_{H_2SO_4}=200.12,25\%=24,5g\\ n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\\ pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\\ LTL:0,2< 0,25=>H_2SO_4\left(d\text{ư}\right)Fe\left(h\text{ết}\right)\) 
    \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\\ V_{H_2}=0,2.22,4=4,48l\)
\(C\%=\dfrac{11,2}{24,5}.100\%=45,7\%\)

`11,2 g` j bạn ?