\(BTNT\left(H\right):n_{H_2SO_4}.2=n_{H_2O}.2\\ \Rightarrow n_{H_2O}=0,6\left(mol\right)\\ BTKL:m_{hh}+m_{H_2SO_4}=m_{muối}+m_{H_2O}\\ \Rightarrow m_{muối}=32+0,6.98-0,6.18=80\left(g\right)\\ \Rightarrow ChọnB\)

\(BTNT\left(H\right):n_{H_2SO_4}.2=n_{H_2O}.2\\ \Rightarrow n_{H_2O}=0,02\left(mol\right)\\ BTKL:m_{hh}+m_{H_2SO_4}=m_{muối}+m_{H_2O}\\ \Rightarrow m_{muối}=2,3+0,02.98-0,02.18=3,9\left(g\right)\\ \Rightarrow ChọnD\)

D

Gọi $n_{Fe} = a(mol) ; n_{FeO} = b(mol) \Rightarrow 56a + 72b = 36,8(1)$
$2Fe + 6H_2SO_4 \to Fe_2(SO_4)_3 + 3SO_2 + 6H_2O$
$2FeO + 4H_2SO_4 \to Fe_2(SO_4)_3 + SO_2 + 4H_2O$

Theo PTHH : 

$n_{SO_2} = 1,5a + 0,5b = 15,68 : 22,4 = 0,7(2)$

Từ (1)(2) suy ra a = 0,4 ; b = 0,2

$n_{Fe_2(SO_4)_3} = (a + b).0,5 = 0,3(mol)$
$m_{Fe_2(SO_4)_3} = 0,3.400 = 120(gam)$

Đáp án: C

Oxit + H₂SO₄→ Muối + H₂O

           0,6 mol                 0,6 mol

BTKL: 32+ 0,6.98=m_muối + 0,6.18

m_muối =80 gam

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\\ m_{H_2SO_4}=0,15.98=14,7\left(g\right)\\ m_{hh\left(Al,Zn,Mg\right)}+m_{H_2SO_4}=m_{hh.muoi}+m_{H_2}\\ \Leftrightarrow5,2+14,7=m+0,15.2\\ \Leftrightarrow m=19,6\left(g\right)\\ \Rightarrow Chon.A\)

Đặt: nCuO=x(mol); nCu=2x(mol) (x>0)

CuO + H2SO4(đ) -to-> CuSO4 + H2O

0,1__0,1___________0,1(mol)

Cu + 2 H2SO4 (đ) -to-> CuSO4 + SO2 + 2 H2O

0,2_____0,4______0,2_________0,2(mol)

V(SO2,đktc)=4,48(l) => nSO2=4,48/22,4=0,2(mol)

=> nCu=0,2(mol) => nCuO= 0,1(mol)

m1= 0,1. 80 + 0,2. 64= 20,8(g)

m2= (0,1+0,2).160=48(g)

=>m1+m2=20,8+48=68,8(g)

=>CHỌN C

Chọn A

Tác dụng vừa đủ nên ta có dạng tổng quát :

O + H₂SO₄ -> SO₄ + H₂O

=> n_O = n_SO4 = 0,5 mol

=> m_muối = m_oxit – m_O + m_SO4 = 68,1g

\(PTHH:\\ Fe+2HCl\to FeCl_2+H_2\\ Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\\ n_{H_2}=\dfrac{1,12}{22,4}=0,05(mol)\\ \Rightarrow n_{Fe}=0,05(mol)\\ \Rightarrow m_{Fe}=0,05.56=2,8(g)\\ \Rightarrow m_{Fe_2O_3}=10,8-2,8=8(g)\)

Chọn B