Câu B chịu nka~~

→nH2O(dd bđ)=1120−190,418=51,6(mol)→nH2O(dd bđ)=1120−190,418=51,6(mol)

$S{}^{}+{}^{}O\to {}^{}S{}^{}$

$\to S{}^{}$ hết, ${}^{}O$ dư

nH2

PTHH: \(SO_3+H_2O\rightarrow H_2SO_4\)

Ta có: \(n_{SO_3}=\dfrac{200}{80}=2,5\left(mol\right)=n_{H_2SO_4}\)

Mặt khác: \(m_{ddH_2SO_4}=1000\cdot1,12=1120\left(g\right)\) \(\Rightarrow m_{H_2SO_4\left(ban.đầu\right)}=1120\cdot17\%=190,4\left(g\right)\)

Bảo toàn khối lượng: \(m_{ddH_2SO_4\left(sau\right)}=m_{SO_3}+m_{ddH_2SO_4\left(ban.đầu\right)}=1320\left(g\right)\)

\(\Rightarrow C\%=\dfrac{2,5\cdot98+190,4}{1320}\cdot100\%\approx32,98\%\)

a) 200ml = 0,2 lít ; n_{\(H_2SO_4\)} = \(\dfrac{29,4}{98}\) = 0,3(mol)

=> C_M = \(\dfrac{0,3}{0,2}\) = 1,5(M)

b) C% = \(\dfrac{C_M.M_{H_2SO_4}}{10D}\) = \(\dfrac{1,5.98}{10.1,2}\) = 12,25%

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

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

Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c, \(n_{H_2SO_4}=n_{Fe}=0,1\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

Giúp mình với mái thì rồi

$2SO_2 + O_2 \xrightarrow{t^o,xt} 2SO_3$

n SO3 = n SO2 = 7,84/22,4= 0,35(mol)

$SO_3 + H_2O \to H_2SO_4$

m dd H2SO4 60% = 57,2.1,5 = 85,8(gam)

Sau khi pha :

m H2SO4 = 85,8.60% + 0,35.98 = 85,78(gam)

m dd = 85,8 + 0,35.80 = 113,8(gam)

C% H2SO4 = 85,78/113,8   .100% = 75,38%

n fe =0.02(mol)  

pthh: fe + h2so4 = feso4 + h2

(mol)0.02 0.02     ->0.02    0.02

=> m feso4 = 0.02 x 152= 3.04 (g)          m dd feso4 = (m fe + m dd h2so4) - m h2

                                                                                    = (1.12 + 200) -(2x 0.02)=201.08(g)

c% feso4 = 1.51%

 bạn nào giúp mình đi chớ giải cả bài mình không hiểu thế vào C% sau PƯ đâu

SO₃ + H₂O \(\rightarrow\) H₂SO₄

n_SO3=\(\dfrac{200}{80}=2,5\left(mol\right)\)

Theo PTHH ta có:

n_SO3=n_H2SO4=2,5(mol)

m_H2SO4=98.2,5=245(g)

m_dd H₂SO₄ 17 % =1000.1,12=1120(g)

m_H2SO4trong dd=1120.\(\dfrac{17}{100}=190,4\left(g\right)\)

C% dd H₂SO₄ =\(\dfrac{245+190,4}{1120+200}.100\%=33\%\)

\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)

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

\(0.1......................0.1..........0.1\)

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{dd_{H_2SO_4}}=150\cdot1.08=162\left(g\right)\)

\(m_{dd}=6.5+162-0.1\cdot2=168.3\left(g\right)\)

\(C\%_{ZnSO_4}=\dfrac{0.1\cdot161}{168.3}\cdot100\%=9.56\%\)

`Fe_2O_3+3H_2SO_4->Fe_2(SO_4)_3+3H_2O`

0,0625----------0,1875---------0,0625 mol

`->n_(Fe_2O_3)=10/160=0,0625mol`

`->m_(Fe_2(SO_4)_3)=0,0625.400=25g`

`->C%(H_2SO_4)=((0,1875.98)/(450)).100%=4,083%`

`#YBtran<3`

\(n_{Fe_2O_3}=\dfrac{10}{160}=0,0625\left(mol\right)\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,0625\left(mol\right)\\ a,m=m_{Fe_2\left(SO_4\right)_3}=400.0,0625=25\left(g\right)\\ b,n_{H_2SO_4}=3.0,0625=0,1875\left(mol\right)\\ C\%_{ddH_2SO_4}=\dfrac{0,1875.98}{450}.100\%\approx4,083\%\)