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k0ka [10]
3 years ago
8

the elements iron and oxygen can react to form the compound iron oxide. what type of reaction is this? please explain.

Chemistry
2 answers:
NemiM [27]3 years ago
8 0

Answer:

Oxidation reaction

Explanation:

Rust is formed when iron reacts with oxygen in moist air. The following chemical equation represents the reaction: 4Fe + 3O2 → 2Fe2O3. Water is necessary for the oxidation reaction to occur and to facilitate transport of the electrons

rjkz [21]3 years ago
4 0
Your reaction
.. Fe + O2 ---> FexOy
for this reaction..
.. the Fe on the left is in the 0 oxidation state
.. the Fe on the right is in the +(2y/x) oxidation state
.. the O on the left is in the 0 oxidation state
.. the O on the right is in the -2 oxidation state
meaning
.. the O is reduced... . . (it's reduced in oxidation state)
.. the Fe is oxidized.. . .(oxidation state increased)
this is a REDOX reaction

*********
AND.. it's also a synthesis reaction.. (aka combination reaction)
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Answer:

0.175\; \rm mol \cdot L^{-1}.

Explanation:

Magnesium chloride and silver nitrate reacts at a 2:1 ratio:

\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s).

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to  \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}.

The precipitate silver chloride \rm AgCl is insoluble in water and barely ionizes. Hence, \rm AgCl\! isn't rewritten as ions.

Net ionic equation:

\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}.

Calculate the initial quantity of nitrate ions in the mixture.

\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}.

Since nitrate ions \rm {NO_3}^{-} do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol.

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be (1/2) of the concentration in the original solution.

\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}.

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