C1 a, MCO=12+16=28g/mol Trong 1 mol hợp chất CO có 1mol ntu C 1mol ntu O +,%mC= 12×100÷28=42,85% +,%mO=100-42,85=57,15% MCO2= 12+16×2=44g/mol Trong 1 mol hợp chất CO2 có 1mol ntu C 2MOL NTU O +,%mC= 12×100÷44=27,3% +,mO= 100-27,3=72,7% b MFe3O4=56×3+16×4=232g/mol Trog 1 mol hc Fe3O4 có 3mol ntu Fe 4mol ntu O +,%mFe= 56×3×100÷232=72,4% +,%mO=100-72,4=27,6% MFe2O3=56×2+16×3=131g/mol Trong 1 mol hc Fe2O3 có 2mol ntu Fe 3mol ntu O +,%mFe=56×2×100÷131=85,5% +,%mO=100-85,5=14,5% MSO2=32+16×2=64g/mol Trong 1 mol hc SO2 có 1mol ntu S 2mol ntu O +,%mS=32×100÷64=50% +,%mO= 100-50%=50% MSO3=32+16×3=80g/mol Trong 1 mol hc SO3 có 1mol ntu S 3mol ntu O +,%mS=32×100÷80=40% +,%mO=100-40=60% Câu 2 MgCl2+2KOH ->Mg(OH)2+2KCl 2Cu(OH)2+2HCl-> 2CuCl2+2H2O Cu(OH)2+H2SO4->CusO4+2H2O FeO+2HCl->FeCl2+H2O Fe2O3+3H2SO4->Fe2(SO4)3+3H2O Câu 3 nFe=8,4÷56=0,15mol nO2=8÷32=0,25mol VO2=0,25×22,4=5,6lit nN2=67,2÷22,4=3mol mN2=3.28=84g Câu 4 VSO2=0,03×22,4=0,672lit VHcl=0,15×22,4=3,36lit VCO2=0,125×22,4=2,4lit VH2SO4=1,2×22,4=26,88lit

Đáp án:vote 5 s

 

Giải thích các bước giải:

a/

+ CO

%m_C = 1212+16.100%=42,86%1212+16.100%=42,86%%m_O = 100% - 42,86% =57,14%​

+CO₂

%m_C = 1212+16.2.100%=27,27%1212+16.2.100%=27,27% %m_O = 100% - 27,27% = 72,73%

b/

+Fe₃O₄

%m_Fe = 56.356.3+16.4.100%=72,41%56.356.3+16.4.100%=72,41% %m_O = 100% - 72,41% = 27,59%

+ Fe₂O₃

%m_Fe = 56.256.2+16.3.100%=70%56.256.2+16.3.100%=70% %m_O = 100% - 70% = 30%

c/

+SO₂

%m_S = 3232+16.2.100%=50%3232+16.2.100%=50%%m_O = 100% - 50% = 50%

+ SO₃

%m_S = 3232+16.3.100%=40%3232+16.3.100%=40% %m_O = 100% - 40% = 60%

câu 2

1) MgCl₂ + KOH → Mg(OH)₂ + KCl
2) Cu(OH)₂ + HCl → CuCl₂ + H₂O
3) Cu(OH)₂ + H₂SO₄ → CuSO₄ + H₂O
4) FeO + HCl → FeCl₂ + H₂O
5) Fe₂O₃ + H₂SO₄ → Fe₂ (SO₄)₃ + H₂O

câu 3

a)nFe=8,4/56=0,15(mol)

b)nO2=8/32=0,25(mol)

=>VO2=0,25.22,4=5,6(l)

c) nN2=67,2/22,4=3(mol)

=>mN2=3.28=84(g)

câu 4