# 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

(會英文的還是看英文的好........歹勢)

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

### 1 個解答

• 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 

n(14+16y)<60 

Solving for y in equation leads to y=0.5, i.e. the ratio of N to O is 2. Replacing y in equation  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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