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222-9+11+12:2*14+14 = ? ( )

Hóa học Lớp 8: 1) CO2 + H2O =-> H2CO3 2) HCL + NaOH ==-> NaCL + H2O 3) MgO + HNO3 ==> Mg(NO3)2 + H2O 4) AL +CuO =–> AL2O3 + Cu 5) AL + HC

Hóa học Lớp 8: 1) CO2 + H2O —-> H2CO3
2) HCL + NaOH ——-> NaCL + H2O
3) MgO + HNO3 ——> Mg(NO3)2 + H2O
4) AL +CuO —–> AL2O3 + Cu
5) AL + HCL ——-> ALCL3 + H2
6) FeO + HCL —-> FeCL2 + H2O
7) Fe2O3 + H2SO4 —–> Fe2(SO4)3 + H2O
8) NaOH + H2SO4 —> Na2SO4 + H2O
9) Zn + HCL —-> ZnCL2 + H2
10) AL + CuCL2 —–> ALCL3 + Cu
11) AL2O3 + HCL —–> ALCL3 + H2O, giúp em giải bài hóa này ạ, em cảm ơn thầy cô và các bạn nhiều.

Comments ( 2 )

  1. Em tham khảo!
    $\text{1) CO2+H2O$\rightarrow$H2CO3}$
    Tỉ lệ: $\text{1:1:1}$
    $\text{2) HCl+NaOH$\rightarrow$NaCl+H2O}$
    Tỉ lệ: $\text{1:1:1:1}$
    $\text{3) MgO+2HNO3$\rightarrow$Mg(NO3)2+H2O}$
    Tỉ lệ: $\text{1:2:1:1}$
    $\text{4) 2Al+3CuO$\xrightarrow{t^o}$Al2O3+3Cu}$
    Tỉ lệ: $\text{2:3:1:3}$
    $\text{5) 2Al+6HCl$\rightarrow$2AlCl3+3H2}$
    Tỉ lệ: $\text{2:6:2:3}$
    $\text{6) FeO+2HCl$\rightarrow$FeCl2+H2O}$
    Tỉ lệ: $\text{1:2:1:1}$
    $\text{7) Fe2O3+3H2SO4$\rightarrow$Fe2(SO4)3+3H2O}$
    Tỉ lệ: $\text{1:3:1:3}$
    $\text{8) 2NaOH+H2SO4$\rightarrow$Na2OS4+2H2O}$ 
    Tỉ lệ: $\text{2:1:1:2}$
    $\text{9) Zn+2HCl$\rightarrow$ZnCl2+H2}$
    Tỉ lệ: $\text{1:2:1:1}$
    $\text{10) 2Al+3CuCl2$\rightarrow$2AlCl3+3Cu}$
    Tỉ lệ: $\text{2:3:2:3}$
    $\text{11) Al2O3+6HCl$\rightarrow$2AlCl3+3H2O}$
    Tỉ lệ: $\text{1:6:2:3}$
     

  2. PTHH:
    (1) CO2 + H2O ⇄ H2CO3
    (2) HCl + NaOH → H2O + NaCl
    (3) MgO + 2HNO3 → Mg(NO3)2 + H2O
    (4) 2Al + 3CuO → Al2O3 + 3Cu↓
    (5) 2Al + 6HCl → 2AlCl3 + 3H2
    (6) FeO + 2HCl → FeCl2 + H2O
    (7) Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O
    (8) 2NaOH + H2SO4 → Na2SO4 + 2H2O
    (9) Zn + 2HCl → ZnCl2 + H2
    (10) 2Al + 3CuCl2 → 2AlCl3 + 3Cu
    (11) Al2O3 + 6HCl → 2AlCl3↓ + 3H2O
     

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222-9+11+12:2*14+14 = ? ( )

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