3.13 g of sodium (Na) react with 7.17 g of chlorine (Cl2) to produce 10.3 g of sodium chloride (NaCl).

What is the mass percent of chlorine in sodium chloride?

50.0% is chlorine.
30.3% is chlorine.
69.6% is chlorine.
43.7% is chlorine.

I do not believe it is 30.3 or 43.7, I believe it is 69.6

FIRST MY WAY

Na = 23 grams / mol
Cl = 35.5 grams/mol
so
NaCl is 23 + 35.5 g/mol = 58.5 g/mol
so in that mol of Na Cl which weighs 58.5 g
you have one mol of Cl which weighs 35.5 g
so
what is 100 * 35.5 / 58.5 = 60.7 %
now what they did (silly)
100 * 7.17/10.3 = 69.6 %

To find the mass percent of chlorine in sodium chloride (NaCl), we can use the formula:

Mass percent of chlorine = (mass of chlorine / mass of sodium chloride) * 100

Given:
Mass of sodium (Na) = 3.13 g
Mass of chlorine (Cl2) = 7.17 g
Mass of sodium chloride (NaCl) = 10.3 g

Now, let's calculate the mass percent of chlorine:

Mass percent of chlorine = (7.17 g / 10.3 g) * 100
= 69.6%

So, the correct answer is:

69.6% is chlorine.

To find the mass percent of chlorine in sodium chloride, we need to calculate the mass of chlorine and the total mass of sodium chloride.

Given:
Mass of sodium (Na) = 3.13 g
Mass of chlorine (Cl2) = 7.17 g
Mass of sodium chloride (NaCl) = 10.3 g

To calculate the mass of chlorine, we need to convert the mass of chlorine gas (Cl2) to the mass of chlorine atoms (Cl). Chlorine is diatomic, meaning it exists as molecules with two chlorine atoms bonded together. So, the molecular weight of chlorine is 35.45 g/mol.

Using the molar mass of chlorine (Cl2), we can calculate the moles of chlorine present:

Moles of chlorine (Cl2) = Mass of chlorine (Cl2) / Molar mass of chlorine (Cl2)
Moles of chlorine (Cl2) = 7.17 g / 70.90 g/mol

From the balanced chemical equation for the synthesis reaction between sodium and chlorine:

2 Na + Cl2 -> 2 NaCl

We can see that one mole of chlorine (Cl2) reacts with two moles of sodium (Na) to produce two moles of sodium chloride (NaCl).

Therefore, the moles of chlorine (Cl) present in the sodium chloride are equal to the moles of chlorine gas (Cl2) used in the reaction:

Moles of chlorine (Cl) = 2 * Moles of chlorine (Cl2)
Moles of chlorine (Cl) = 2 * (7.17 g / 70.90 g/mol)

Now, we can calculate the mass of chlorine (Cl) present in the sodium chloride:

Mass of chlorine (Cl) = Moles of chlorine (Cl) * Molar mass of chlorine (Cl)
Mass of chlorine (Cl) = (2 * Moles of chlorine (Cl2)) * 35.45 g/mol

Finally, we can calculate the mass percent of chlorine in sodium chloride:

Mass percent of chlorine in sodium chloride = (Mass of chlorine (Cl) / Mass of sodium chloride (NaCl)) * 100

Let's calculate it:

Mass percent of chlorine in sodium chloride = (Mass of chlorine (Cl) / 10.3 g) * 100

Using the above steps, the calculated mass percent of chlorine in sodium chloride should be 69.6%. Therefore, your belief is correct.

Not only is it silly; it is not possible for those numbers.

Starting with 3.13 g Na you should, at 100% yield, obtain *2*58.5/2*23 = 7.96 g NaCl but they said 10.3. Can't be, that's better than 100% yield. OR,
starting with 7.17 g Cl2 you would get, at 100% yield, 7.17 x (2*58.5/71) = 11.8 g. Can't be unless the reaction was < 100%. Dead cat on the line somewhere. 60.7% chlorine in NaCl is correct.(71/2*58.5 )100 = 60.68% which rounds to 60.7%. OR,
The limiting reagent is Na @ 3.13 g. How much Cl2 is needed to react @ 100%. That is 3.13*71/2*23 = 4.83.(obviously there is excess Cl2)
Total NaCl is g Na + g Cl2 = 3.13 + 4.83 = 7.96. Now the coup de grace is
%Chlorine = (g chlorine/g NaCl total)100 = 100 x 4.83/7.95 = 60.68 or 60.7%. Voila!