S+O2->SO2

nSO2=0.4(mol)

Theo pthh nS=nO2=nSO2->nS=nO2=nSO2=0.4(mol)

m1=0.4*32=12.8(g)

m2=0.4*32=12.8(g)

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

\(2Zn+O_2\underrightarrow{^{^{t^0}}}2ZnO\)

\(0.1.......0.05.......0.1\)

\(V_{O_2}=0.05\cdot22.4=1.12\left(l\right)\)

\(m_{ZnO}=0.1\cdot81=8.1\left(g\right)\)

C₂H₂ + \(\frac{5}{2}\)O₂ => (t^o) 2CO₂ + H₂O

0.25 0.625 0.5 0.25 (mol)

n_CO2 = V/22.4 = 11.2/22.4 = 0.5 (mol);

m₁ = n.M = 0.25 x 26 = 6.5 (g);

m₂ = n.M = 0.25 x 18 = 4.5 (g);

m₁ + m₂ = 6.5 + 4.5 = 11 (g);

V = VO2 = n.22.4 = 0.625x22.4=14(l)

a) \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: 3Fe + 2O₂ --t^o--> Fe₃O₄

           0,3<---0,2-------->0,1

=> m = 0,3.56 = 16,8 (g)

b) m_Fe3O4 = 0,1.232 = 23,2 (g)

c) V_kk = 4,48 : 20% = 22,4 (l)

nO2 = 4,48/22,4 = 0,2 (mol)

PTHH: 3Fe + 2O2 -> (t°) Fe3O4

Mol: 0,3 <--- 0,2 ---> 0,1

mFe = 0,3 . 56 = 16,8 (g)

mFe3O4 = 0,1 . 232 = 23,2 (g)

Vkk = 4,48 . 5 = 22,4 (l)

\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: 3Fe + 2O₂ --t^o--> Fe₃O₄

           0,6<--0,4

=> m_Fe = 0,6.56 = 33,6(g)

\(n_{O_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)

\(0.6......0.4\)

\(m_{Fe}=0.6\cdot56=33.6\left(g\right)\)

\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

\(PTHH:3Fe+2O_2->Fe_3O_4\)

                0,13     0,2           0,4

\(m_{Fe}=0,13.56=7,28\left(g\right)\)

Gọi CTPT của Z là $C_nH_{2n-4}O_2$

$n_{O_2} = 0,4(mol) ; n_{CO_2} = 0,4(mol)$

Bảo toàn electron  :$n_Z.(4n + 2n - 4 - 2.2) = 4n_{O_2}$

Bảo toàn C : $n_Z.n = n_{CO_2}$

Suy ra: 

$\dfrac{0,4.4}{6n-8} = \dfrac{0,4}{n} \Rightarrow n = 4$

Suy ra: Z là $C_4H_4O_2$

$n_Z = \dfrac{0,4}{4} = 0,1(mol)$

Bảo toàn H : $n_{H_2O} = 2n_Z = 0,2(mol)$

Suy ra : $m_1 + m_2 = 0,1.84 + 0,2.18 = 12(gam)$

Chọn A

Chọn A