Limiting Reagent Reaction Question

[Lyssaaa]

Hi! So I was given this question to do and I'm not sure of what the answers are as we haven't been given them. I completed the question but I don't know if I did it quite right. Any solutions would be amazing!

Sodium metal can react with chlorine gas to form sodium chloride. 25.0g of sodium is reacted with 50.0g of chlorine gas. Calculate the mass of sodium chloride that will form in the reaction.

Thank you!

[fun_jirachi]

Welcome to the forums!!

The first step should always be to write a chemical equation:
Na_(s)+Cl_(g) --> NaCl_(s)

Now, 25.0g of sodium is equal to 1.08742932 moles, while 50.0g of chlorine is equal to 1.41043724 moles. Notice that from the chemical equation that you require sodium and chlorine in a 1:1 ratio, and hence from the number of moles of each chemical we have, we have excess chlorine. NaCl is also in a 1:1 ratio with each of the reactants, so we can actually skip a step and say that the number of moles of NaCl is equal to the number of moles of reacting sodium and chlorine, which is 1.08742932. Multiply this number by the molar mass of sodium chloride, and you should get the mass of NaCl that will form, which should be (if I'm not wrong) 63.5g (3SF).

Hope this helps

[Lyssaaa]

Thank you very much! As it specifies that chlorine is a gas in this reaction, would I need to balance the equation as Cl2 and so forth? In that case, would the answer still be the same? I got the same mass as you had calculated but just wanted to double check!

Thanks again!

[r1ckworthy]

As it specifies that chlorine is a gas in this reaction, would I need to balance the equation as Cl2 and so forth? In that case, would the answer still be the same? I got the same mass as you had calculated but just wanted to double check!

I don't think this will be much of an issue, because even when you use Cl2, you have to change NaCl from 1 to 2, and then change Na to 2 in order to balance the reaction. This will lead to all the chemical species having a molar ratio of 2 (2:2:2) which simplifies to 1 (1:1:1).
           
EDIT: Very silly mistake I made, just follow fun_jirachi. I assumed Cl2 will have a molar ratio of 2 which is really bad of me 

[fun_jirachi]

Thank you very much! As it specifies that chlorine is a gas in this reaction, would I need to balance the equation as Cl2 and so forth? In that case, would the answer still be the same? I got the same mass as you had calculated but just wanted to double check!

Thanks again!

Thanks for pointing that out! That is a mistake
You're correct there actually; the equation should in fact be
2Na_(s)+Cl_2(g)-->2NaCl_(s)
Then, in this case, you'd still have 1.08742932 moles of sodium, but 0.70521862 moles of chlorine. Hence, you'd produce 1.08742932 moles of NaCl since the ratio of sodium to sodium chloride is 1:1, and since you need two moles of sodium for one mole of chlorine, and as such you'd produce 63.5g of sodium chloride.

Sorry for the mistake! Hope this helps

[Lyssaaa]

Okay great! Thank you so much for clarifying!