\(n_{Ba}=\frac{41,1}{137}=0,3\left(mol\right)\)

\(n_{H2SO4}=\frac{4,9\%.200}{98}=0,1\left(mol\right)\)

\(\text{Ba+2H2O}\rightarrow\text{Ba(OH)2+H2}\)

0,3____________0,3_____0,3

\(\text{Ba(OH)2+H2SO4}\rightarrow\text{BaSO4+2H2O}\)

0,3_______0,1_______0,1

\(\rightarrow\text{VH2=0,3.22,4=6,72l}\)

b) Vdd=200-0,3.2-0,1.223=176,1g

nBa(OH)2 dư= 0,3-0,1=0,2 mol

\(\rightarrow\)m Ba(OH)2 dư= 0,2.171=34,2g

\(\rightarrow\)\(C\%=\frac{\text{34,2}}{\text{176,1}}.100\%\text{=19,42%}\)

a) ba+2H2O---->Ba(OH)2+H2O

Ba(OH)2+H2SO4---->BaSO4+2H2O

n Ba=41,4/137=0,3(mol)

Theo pthh1

n H2=n Ba=0,3(mol)

V H2=0,3.22,4=6,72(l)

b)

Theo pthh1

n Ba(OH)2=n Ba=0,3(mol)

n H2SO4=\(\frac{200,4,9}{100.98}=0,1\left(mol\right)\)

--->Ba(OH)2 dư

Dd sau pư là Ba(OH)2 dư

Theo pthh2

n Ba(OH)2=n H2SO4=0,1(mol)

--->n Ba(OH)2 dư=0,2(mol)

, dd=200+0,2.171=234,2(g)

C%=\(\frac{0,2.171}{234,2}.100\%=14,6\%\)

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

`11,2 g` j bạn ?

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

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%.

Zn+H2SO4->ZnSO4+H2

0,05-------------------------0,05

n Zn=\(\dfrac{4}{65}\) mol

n H2SO4=\(\dfrac{4,9}{98}\)=0,05 mol

=>Zn dư

m Zn=(\(\dfrac{4}{65}\)-0,05).65=0,75 g

VH2=0,05.22,4=1,12l

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

Ta có: \(n_{Ba}=\dfrac{8,22}{137}=0,06\left(mol\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{m_{H_2SO_4}}{200}.100\%=1,96\%\)

=> \(m_{H_2SO_4}=3,92\left(g\right)\)

=> \(n_{H_2SO_4}=\dfrac{3,92}{98}=0,04\left(mol\right)\)

PTHH: Ba + H₂SO₄ ---> BaSO₄↓ + H₂

Ta thấy: \(\dfrac{0,06}{1}>\dfrac{0,04}{1}\)

=> Ba dư

Theo PT: \(n_{BaSO_4}=n_{H_2SO_4}=0,04\left(mol\right)\)

=> \(m_{BaSO_4}=0,04.233=9,32\left(g\right)\)

Theo PT: \(n_{H_2}=n_{H_2SO_4}=0,04\left(mol\right)\)

=> \(m_{H_2}=0,04.2=0,08\left(g\right)\)

Ta có: \(m_{dd_{BaSO_4}}=8,22+200-0,08=208,14\left(g\right)\)

=> \(C_{\%_{BaSO_4}}=\dfrac{9,32}{208,14}.100\%\approx4,48\%\)

Coi như p/ứ vừa đủ

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3 \left(mol\right)=n_{H_2SO_4}=n_{ZnSO_4}=n_{H_2}\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4}=\dfrac{0,3\cdot98}{200}=14,7\%\\V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\\m_{ZnSO_4}=0,3\cdot161=48,3\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{Zn}+m_{ddH_2SO_4}-m_{H_2}=218,9\left(g\right)\)

\(\Rightarrow C\%_{ZnSO_4}=\dfrac{48,3}{218,9}\cdot100\%\approx22,06\%\)

nNa = 4.6/23 = 0.2 (mol) 

Na + H2O => NaOH + 1/2H2

0.2....................0.2..........0.1

VH2 = 0.1*22.4 = 2.24 (l) 

mNaOH = 0.2*40 = 8 (g) 

Đề thiếu khối lượng nước rồi em nhé !

\(1) 2Na + 2H_2O \to 2NaOH + H_2\\ 2) n_{Na} = \dfrac{4,6}{23} = 0,2(mol)\\ n_{H_2} = \dfrac{1}{2}n_{H_2} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ 3) n_{NaOH} = n_{Na} = 0,2(mol)\\ m_{NaOH} = 0,2.40 = 8(gam)\)

4) Thiếu dữ kiện

Đề đâu em nhỉ ?