谁帮我看看这个化学方程式呀Ni(NO3)2 + 3N2H4.H2O = {Ni(N2H4)3}(NO3)2 + 3H2O

1个回答

  • 分子量:

    M :Ni(NO3)2

    58.69 + (14.01+16.00*3)*2 = 182.71g/mol

    M :{Ni(N2H4)3}(NO3)2

    58.69 + (14.01*2+1.01*4)*3 + (14.01+16.00*3)*2

    =58.69+96.18+124.02

    =278.89g/mol

    M :N2H4.H2O

    14.01*2+1.01*4+18.02=50.08g/mol

    摩尔系数比:

    Ni(NO3)2 + 3N2H4.H2O = {Ni(N2H4)3}(NO3)2 + 3H2O

    1 3 1 3

    摩尔数是:

    根据得率按80%计

    {Ni(N2H4)3}(NO3)2的摩尔数是:

    (30/80%)/278.89mol=0.1345mol

    则 Ni(NO3)2 消耗的摩尔数是:0.1345mol

    则 N2H4.H2O 消耗的摩尔数是:3*0.1345mol

    质量:

    Ni(NO3)2 :

    0.1345mol*182.71=24.5745g

    N2H4.H2O:(根据其为浓度40%)

    3*0.1345mol*50.08/40%=50.5182g

    不好意思刚才算错了,公家的饭不等人哈,久等了!