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Nonamiya [84]
3 years ago
12

When magnesium metal is heated in air it begins to release large amounts of heat and light. What kind of a reaction is this an e

xample of?
Chemistry
2 answers:
Salsk061 [2.6K]3 years ago
5 0
It is called exothermic reaction because it releases heat and light and it is called combustion reaction because it is reacting and is being oxidised by O2 to MgO. It can also be called as oxidation reaction since Mg is oxidised to MgO.
Ipatiy [6.2K]3 years ago
4 0
If we're talking about endothermic and exothermic reactions, then it is an exothermic reaction, since heat is being released.
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2 years ago
If a sample containing 18.1 g of NH3 is reacted with 90.4 g of
USPshnik [31]

Answer:

3.64g

Explanation:

Given parameters:

Mass of NH₃  = 18.1g

Mass of Cu₂O  = 90.4g

Unknown:

Limiting reactant  = ?

Mass of N₂ formed  = ?

Solution:

The reaction equation is given as:

       Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O

The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;

  Number of moles = \frac{mass}{molar mass}  

Molar mass of Cu₂O = 2(63.6) + 16  = 143.2g/mol

Molar mass of NH₃  = 14 + 3(1) = 17g/mol

Number of moles of Cu₂O = \frac{18.1}{143.2}   = 0.13moles

Number of moles of NH₃   = \frac{90.4}{17}   = 5.32moles

  From this reaction;

       1 mole of  Cu₂O combines with 2 mole of NH₃

So   0.13moles of  Cu₂O will combine with 0.13 x 2 mole of NH₃

                                              = 0.26moles of NH₃

Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;

Mass of N₂;

   Mass = number of moles x molar mass

    1 mole of Cu₂O  will produce 1 mole of N₂

    0.13 mole of Cu₂O  will produce 0.13 mole of N₂

    Mass  = 0.13 x (2 x 14) = 3.64g

5 0
3 years ago
Under the right conditions aluminum will react with chlorine to produce aluminum chloride.
salantis [7]

Answer:

m_{Al}=9.42gAl

Explanation:

Hello there!

In this case, according to the given chemical reaction:

2 Al + 3 Cl2 --> 2 AlCl3

Whereas there is a 2:3 mole ratio of aluminum to chlorine; it will be possible for us to calculate the required grams of aluminum by using the equality 22.4 L = 1 mol, the aforementioned mole ratio and the atomic mass of aluminum (27.0 g/mol) to obtain:

m_{Al}=11.727LCl_2*\frac{1molCl_2}{22.4LCl_2}*\frac{2molAl}{3molCl_2}  *\frac{27.0gAl}{1molAl} \\\\m_{Al}=9.42gAl

Regards!

8 0
2 years ago
What needs to happen at point 3 for igneous rock to transform into sedimentary rock?
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