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vazorg [7]
3 years ago
15

Sodium is a soft, white metal that reacts violently with water. Chlorine is a poisonous yellow-green gas. During the chemical re

action between sodium and chlorine, table salt is produced. Which of the following statements correctly compares and contrasts the reactants with the resulting products used in a chemical equation?
A. The reactants are dangerous and the product can be eaten
B. The reactants and the products observe the law of conservation of mass
C. The reactants are a solid and a gas and the product is a gas
D. The reactants and the products are the same color
Chemistry
1 answer:
kompoz [17]3 years ago
6 0

Answer:

B. The reactants and the products observe the law of conservation of mass

Explanation:

The reaction equation is given as:

            2Na  +  Cl₂  →   2NaCl

Such a chemical reactions usually obeys the law of conservation of mass. The law states that "in a chemical reaction, matter is neither created nor destroyed but bonds are rearranged".

  • So, the most correct thing is that reactants and products observe the law of conservation of mass.
  • The number of atoms on both sides of the expression will be the same.
  • Therefore, option B is correct.  
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A 100 g sample of potassium chlorate, KCIO3(s), is completely decomposed by heating:
Mama L [17]
Explanation:
In order to be able to calculate the volume of oxygen gas produced by this reaction, you need to know the conditions for pressure and temperature.
Since no mention of those conditions was made, I'll assume that the reaction takes place at STP, Standard Temperature and Pressure.
STP conditions are defined as a pressure of
100 kPa
and a temperature of
0
∘
C
. Under these conditions for pressure and temperature, one mole of any ideal gas occupies
22.7 L
- this is known as the molar volume of a gas at STP.
So, in order to find the volume of oxygen gas at STP, you need to know how many moles of oxygen are produced by this reaction.
The balanced chemical equation for this decomposition reaction looks like this
2
KClO
3(s]
heat
×
−−−→
2
KCl
(s]
+
3
O
2(g]
↑
⏐
⏐
Notice that you have a
2
:
3
mole ratio between potassium chlorate and oxygen gas.
This tells you that the reaction will always produce
3
2
times more moles of oxygen gas than the number of moles of potassium chlorate that underwent decomposition.
Use potassium chlorate's molar mass to determine how many moles you have in that
231-g
sample
231
g
⋅
1 mole KClO
3
122.55
g
=
1.885 moles KClO
3
Use the aforementioned mole ratio to determine how many moles of oxygen would be produced from this many moles of potassium chlorate
1.885
moles KClO
3
⋅
3
moles O
2
2
moles KClO
3
=
2.8275 moles O
2
So, what volume would this many moles occupy at STP?
2.8275
moles
⋅
22.7 L
1
mol
=
64.2 L
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