2Na + Cl2 - ?NaCI
In the reaction, sodium and chlorine combine to produce sodium chloride. In an instance when eight sodium chloride molecules
are produced, how many diatomic chlorine molecules went into the reaction?
you have 2 Na on the left, so you need 2 Na on the right.
2Na + Cl2 = 2NaCl
Now check the Cl atoms. OK as is: 2 on both sides.
The equation tells you you need 1/2 as many Cl2 as you have NaCl.
So, 4 Cl2 to produce 8 NaCl.
To determine the number of diatomic chlorine (Cl2) molecules that went into the reaction, we can use the stoichiometry of the balanced equation.
The balanced equation for the reaction is:
2 Na + Cl2 -> 2 NaCl
From the equation, we can see that 2 moles of sodium (2 Na) react with 1 mole of diatomic chlorine (Cl2) to produce 2 moles of sodium chloride (2 NaCl).
Since you want to know the number of diatomic chlorine molecules that produced 8 sodium chloride molecules, we need to convert the given information to moles.
One mole of a substance contains Avogadro's number of particles, which is about 6.022 x 10^23 particles.
So, if 2 moles of sodium chloride (2 NaCl) are produced, this corresponds to 1 mole of diatomic chlorine (Cl2) being consumed.
If 1 mole of Cl2 = 6.022 x 10^23 molecules, then 2 moles of Cl2 will contain:
2 moles Cl2 = 2 x (6.022 x 10^23) molecules Cl2 = 1.2044 x 10^24 molecules Cl2
Therefore, in the given instance, when 8 sodium chloride molecules are produced, the reaction consumed:
(8/2) x 1.2044 x 10^24 molecules Cl2 = 4.8176 x 10^24 molecules Cl2
So, 4.8176 x 10^24 diatomic chlorine molecules went into the reaction.
To determine the number of diatomic chlorine molecules that went into the reaction, we need to analyze the ratio of sodium to chlorine in the balanced equation.
The balanced chemical equation for the reaction is:
2Na + Cl2 -> 2NaCl
From this equation, we can see that 2 moles of sodium (2Na) react with 1 mole of diatomic chlorine (Cl2) to produce 2 moles of sodium chloride (2NaCl).
Since the equation states that for every 2 moles of sodium chloride produced, 1 mole of diatomic chlorine is needed, we can set up a simple proportion:
2NaCl / 1Cl2 = 8NaCl / X
Where X is the unknown number of diatomic chlorine molecules.
Simplifying the proportion, we find:
2 / 1 = 8 / X
Cross-multiplying, we get:
2X = 8 * 1
2X = 8
X = 8 / 2
X = 4
Therefore, 4 diatomic chlorine molecules went into the reaction to produce 8 sodium chloride molecules.