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

Hóa học Lớp 8: : Tính khối lượng mol của các chất sau: Fe, Zn, H2, O2, N2, CO, CO2, Na2O, NH3, AlCl3, HCl, H2SO4, K3PO4

Hóa học Lớp 8: : Tính khối lượng mol của các chất sau: Fe, Zn, H2, O2, N2, CO, CO2, Na2O, NH3, AlCl3, HCl, H2SO4, K3PO4, giúp em giải bài hóa này ạ, em cảm ơn thầy cô và các bạn nhiều.

Comments ( 2 )

  1. $1)4K+O_22K_2O$
    $4:1:2$
    $2)2Cu+O_2\xrightarrow{t^o}2CuO$
    $2:1:2$
    $3)2HgO\xrightarrow{đpdd}2Hg+O_2$
    $2:2:1$
    $4)4Al+3O_2\xrightarrow{t^o}2Al_2O_3$
    $4:3:2$
    $5)2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O$
    $2:1:3$
    $6)2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3$
    $2:3:2$
    $7)BaCl_2+Na_2SO_4\to BaSO_4\downarrow+2NaCl$
    $1:1:1:2$
    $8)FeO+2HCl\to FeCl_2+H_2O$
    $1:2:1:1$
    $9)2SO_2+O_2\xrightarrow[V_2O_5]{t^o}2SO_3$
    $2:1:2$
    $10)2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2$
    $2:2:3$
    $11)3Ca(OH)_2+2FeCl_3\to 3CaCl_2+2Fe(OH)_3\downarrow$
    $3:2:3:2$
    $12)2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$
    $2:1:1:2$
    $13)2Al+6HCl\to 2AlCl_3+3H_2$
    $2:6:2:3$
    $14)Fe(OH)_3+3HCl\to FeCl_3+3H_2O$
    $1:3:1:3$
    $15)CaCl_2+2AgNO_3\to Ca(NO_3)_2+2AgCl\downarrow$
    $1:2:1:2$
    $16)3Mg+2AlCl_3\to 3MgCl_2+2Al$
    $3:2:3:2$
    $17)2Al+3CuCl_2\to 2AlCl_3+3Cu\downarrow$
    $2:3:2:3$
    $18)3Fe+2H_3PO_4\to Fe_3(PO_4)_2+3H_2$
    $3:2:1:3$
    $19)BaCl_2+K_2SO_4\to BaSO_4\downarrow+2KCl$
    $1:1:1:2$
    $20)2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$
    $2:1:1:1$

  2. M_{Fe} = 56 $g/mol$
    M_{Zn} = 65 $g/mol$
    M_{H_2} = 2 . 1 = 2 $g/mol$
    M_{O_2} = 16 . 2 = 32 $g/mol$
    M_{N_2} = 14 . 2 = 28 $g/mol$
    M_{CO} = 12 + 16 = 28 $g/mol$
    M_{CO_2} = 12 + 32 = 44 $g/mol$
    M_{Na_2O} = 23 . 2 + 16 = 62 $g/mol$
    M_{NH_3} = 3 + 14 = 17 $g/mol$
    M_{AlCl_3} = 27 + (35,5 . 3) = 133,5 $g/mol$
    M_{HCl} = 1 + 35,5 = 36,5 $g/mol$
    M_{H_2SO_4} = 2 + 32 + 16 . 4 = 98 $g/mol$
    M_{K_3PO_4} = 39 . 3 + 31 + 16 . 4 = 212 $g/mol$

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