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

b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)

c, \(n_{H_2SO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{FeSO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow m_{FeSO_4}=0,35.152=53,2\left(g\right)\)

e, \(C_{M_{FeSO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{H_2SO_4}=0,25.1,6=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{n_{Fe}}{1}< \dfrac{n_{H_2SO_4}}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)

a) theo bài ra ta có :

m_MgO = 4g => n_MgO = \(\dfrac{4}{40}=0,1\left(mol\right)\)

m_H2SO4 = \(\dfrac{200.19,6}{100}=39,2\left(g\right)\)

=> n_H2SO4 = \(\dfrac{39,2}{98}=0,4mol\)

pthh:

MgO + H2SO4 -> MgSO4 + H2O

0,1mol....0,4mol

theo pt : n_MgO = n_H2SO4

theo gt : n_MgO < n_H2SO4

=> H2SO4 dư 0,3mol

=> m_H2SO4(pư) = 0,1 mol

=> m_H2SO4(pư) = 0,1.98 = 9,8 g

vậy có 9,8g axit tham gia pư

b) m_{ddsau pư =} 4 + 200 = 204g

theo pt : n_MgSO4 = 0,1 mol

=> m_MgSO4 = 0,1 . 120 = 12g

=> %m_MgSO4 = \(\dfrac{12.100}{204}=5,88\%\)

ta có n_H2SO4(dư) = 0,3mol

=> m_H2SO4(dư) = 0,3.98 = 29,4g

=>%m_H2SO4(dư) = \(\dfrac{29,4.100}{204}=14,41\%\)

$MgO+ 2HNO_3 \to Mg(NO_3)_2 + H_2O$
$m_{dd\ sau\ pư} =6 + 200 = 206(gam)$
$n_{Mg(NO_3)_2} = n_{MgO} = \dfrac{6}{40} = 0,15(mol)$
$C\%_{Mg(NO_3)_2} = \dfrac{0,15.148}{206}.100\% = 10,78\%$

\(n_{Mg}=\dfrac{10,8}{24}=0,45\left(mol\right)\\ pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\) 
           0,45     0,45                          0,45 
\(m_{H_2SO_4}=0,45.98=44,1\left(g\right)\\ C\%_{H_2SO_4}=\dfrac{44,1}{176,4}.100\%=25\%\\ V_{H_2}=0,45.22,4=10,08\left(l\right)\)

1.pt Fe₂O₃+ 6HCl===>  2FeCl₃+3H₂0

       0,05===>0,3             0,25

==> m muối thu được= 0,25.*( 56+ 35,5.3)=40,625(g)

2. ta có

C_M= n/v => v=n/ Cm=0,6(l)

=> Cm( Fe203)= 0,05/(0.6)=1/12

a) mCuO= 3,2/80= 0,04(mol)

mH2SO4= 40%.100=40(g)

=>nH2SO4=40/98=20/49(mol)

PTHH: CuO + H2SO4 -> CuSO4 + H2O

Ta có: 0,04/1 < 20/49:1

=> H2SO4 dư, CuO hết -> Tính theo nCuO

=> nH2SO4(P.Ứ)=nCuSO4=nCuO=0,04(mol)

=>mH2SO4(p.ứ)=0,04.98=3,92(g)

b) mCuSO4=0,04.160=6,4(g)

c) mH2SO4(dư)= 40 - 3,92= 36,08(g)

mddsau= 3,2+100=103,2(g)

=>C%ddH2SO4(dư sau p.ứ)= (36,08/103,2).100=34,961%

Đáp án A

ý a bạn nhé

ý b bạn nhé

\(1.SO_3+H_2O\rightarrow H_2SO_4\\ 2.m_{H_2SO_4}=\dfrac{200.19,6\%}{100\%}=39,2g\\ n_{H_2SO_4}=\dfrac{39,2}{98}=0,4mol\\ n_{SO_2}=n_{H_2SO_4}=0,4mol\\ m=m_{SO_2}=0,4.64=25,6g\)