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MIG 發問時間: 科學化學 · 1 0 年前

一個算化學成分的問題= =

an iron nail reacts with nitric acid producing iron(II)nitrate, water and an

oxide of nitrogen the oxide of nitrogen is 63.6% nitrongen. and has a molar mass less than 60g/mol. what mass of iron (II)nitrate can be produced from 1.00g of iron and 1.00g of nitric acid

這個本來是英文問題拉.....

但是為了大家方便閱讀所以勉勉強強翻成中文

(會英文的還是看英文的好........歹勢)

一跟鐵丁與nitric acid(沒記錯的話化學式是NO2)反應會生成

氮化鐵(II)、(Fe(NO2)2)水、跟oxide of nitrogen。

oxide of nitrogen佔用了63.6%的淡 而且少於60g/mol的molar mass。

那麼一公克的鐵跟一公克的nitric acid能生成多少的氮化鐵。

這題我解不出來的原因是似乎要解出oxid of nitrogen佔有的百分比跟它的化學式才有辦法......我就是死在這裡= =

似乎不理會氮的數量直接用鐵來換算的話是不行的.....

麻煩看的懂的又能解題的大大了....

1 個解答

評分
  • ?
    Lv 7
    1 0 年前
    最佳解答

    Assume the reaction to be

    Fe + 2xHNO3 --> Fe(NO3)2 + xH2O + (2x-2)NOy

    There are 63.6% N in NOy and the molar mass of NOy is less than 60. Therefore,

    14/(14+16y)=0.636 [1]

    n(14+16y)<60 [2]

    Solving for y in equation leads to y=0.5, i.e. the ratio of N to O is 2. Replacing y in equation [2] results in n<3. The maximum number of n is 2 which tells us that the oxide of nitrogen is N2O. Therefore, the reaction can be rewritten as

    4Fe + 10HNO3 --> 4Fe(NO3)2 + 5H2O + N2O

    1 g Fe contains 1/55.85=0.0179 moles while 1 g nitric acid contains 1/63=0.0159 moles. The ratio of iron to nitric acid is less than 4 to 10. Therefore nitric acid is the limiting reactant. Four moles of iron(II) nitrate are produced per ten moles of nitric acid used. Therefore, the amount of iron(II) nitrate produced is 0.0159*4*179.85/10=1.14 g.

    此題必須先解出氧化氮的化學式才能平衡反應式,找出限量反應物,從而求出生成物產量.

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