if i mix 11.0 gram of sodium with 17.0 gram of chlorine, what mass of sodium chloride would i expect, and how?

Question

do you know?

Answer 1

In order to solve this problem you need to find the limiting reagent in the reaction...either Sodium or Chlorine.

Write out the chemical reaction,
2 Na + Cl2 --> 2 NaCl
You can find a mole ration between the number of moles of Na to Cl2 required for the reaction to use up all of its reactants. If one reactant is in excess, that reactant remains after the reaction is complete and does not form the product.

In the reaction between Na and Cl2, the mole ratio is 2 moles of Na to 1 mole of Cl2.

Now convert the masses of each reactant you are given in the problem into moles of the reactant by dividing by their molar masses.
moles of Na = 11.0 g Na / 22.99 g/mol Na
moles of Na = .478 moles

moles of Cl2 = 17.0 g / 70.9 g/mol Cl2
moles of Cl2 = .240 moles

Now, using the mole ratio which you have already found (2 Na :1 Cl2), determine the limiting reactant.
It turns out, however, that both of these substances are mixed very well and there is no limiting react (or only a negligible amount which I think we can ignore...unless that is indeed the point of the question, but I don’t think it is).

We know there are about .240 moles of Cl2 which react.
We also know a mole ratio which relates the moles of Cl2 which react to the number of moles of NaCl which are formed....1:2. (or we could just as easily use the ratio of Na reacted to NaCl formed, it does not matter)

Multiplying .240 moles by 2 to find the number of moles of NaCl formed we find that there are .480 moles of NaCl formed.
Now that we know the number of moles of NaCl, we can multiply by the molar mass to find the mass of the NaCl formed.

4.80 moles of NaCl * 58.44 g/mol NaCl = 28 grams of NaCl.
You will notice that 28 grams of NaCl equals the sum of the masses of each of the reactants (11 g + 17 g = 28 g), thus mass is conserved.

Answer 2

First write the reaction:
2Na + Cl2 ---> 2NaCl.

Now calculate the moles of sodium ( and chlorine you have.
Find their atomic/molecular weights using the periodic table.

Na, Sodium: 23
Cl2, Chlorine: 2 * 35.5 = 71

Now calculate the number of moles by dividing the mass by the atomic or molecular weight.

Na: 11/23 = 0.48
Cl2: 17/71 = 0.24

From the reaction you see that 2 moles of Na combine with one mole of Cl2. So 0.48 moles of sodium will require 0.48/2 = 0.24 moles of Cl2. You are lucky, you have exactly what is needed.

So you can see that 2 moles of Na will produce 2 moles of NaCl.
So 0.48 moles of Na will produce 0.48 moles of NaCl.

Now you multiply the moles of NaCl by the molecular weight of NaCl (58.5) to obtain the amount of grams.

0.48 mol * 58.5 g/mol = 28.1 g NaCl


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Answer 3

11.0g Na * 1mole/22.99g (the molecular weight)=0.478 moles Na
17.0g Cl2 * 1mole/70.90g (the molecular weight*2 because chlorine is diatomic)=0.239 moles Cl2

2 Na + 1 Cl2 = 2 NaCl

Since they react 2:1 and there is twice as many moles of Na this means that all of the reactants will be used up (nothing in excess).

Since it is all used up you know you will have 28.0 grams (law of conservation of mass). To check our work we cna also calculate it from the moles Nacl formed.

0.478 moles NaCl * 58.44g/mole (the molecular weight of NaCl)=28.0 grams

Answer 4

Na + Cl --> NaCl
well first you figure out which is the Limiting Reactant.
11g Na* (1 mol Na/22.99 g Na)* (1 mol NaCL/1 mole Na) = .4785 mol NaCl
17g Cl * (1 mol Cl/ 35.453 g Cl )* (1 mol NaCl/1molCl)=.4795 mol NaCl
therefore Na is the Limiting Reactant because it produces less NaCl.
now you just take the moles of NaCl produced (.4785 mol) and change into grams
the final answer is 27.965 g NaCl