Question

For the reaction 2NaI + Cl2 →2NaCl + I2 How many grams of NaCl is obtained if 5.0 grams of NaI reacts with 5.0 grams of Cl2?

Answer 1

`0.97528 \ "g"~~1 \ "g"`

Explanation:

We have the balanced equation (without state symbols), as

`2NaI+Cl_2->2NaCl+I_2`

What I first try to do is to find the limiting reactant with these types of problems, so let's do that first.

`NaI` has a molar mass of `149.89 \ "g/mol"`, so `2NaI` has a molar mass of `149.89*2=299.78 \ "g/mol"`.

An average `Cl` atom has a molar mass of `35.453 \ "g/mol"`, so the average molar mass of `Cl_2` will be `35.453*2=70.906 \ "g/mol"`.

The mole ratio between `NaI` to `Cl_2` is `2:1`, so two moles of sodium iodide are needed to react with one mole of chlorine.

We have here `5` grams of `NaI` and also `5` grams of `Cl_2`, and let's convert those amounts into moles.

`"mol"_(NaI)=(5color(red)cancelcolor(black)"g")/(299.78color(red)cancelcolor(black)"g""/mol")~~1.67*10^-2 \ "mol"`

`"mol"_(Cl_2)=(5color(red)cancelcolor(black)"g")/(70.906color(red)cancelcolor(black)"g""/mol")=7.05*10^-2 \ "mol"`

Since we have `7.05*10^-2 \ "mol of" \ Cl_2`, then we would need `7.05*10^-2*2=1.41*10^-1 \ "mol of" \ NaI`.

We have only `1.67*10^-2 \ "mol of" \ NaI`, and `1.67*10^-2<1.41*10^-1`, and so `NaI` becomes the limiting reactant.

So, only `1.67*10^-2` moles of sodium iodide will react fully until there is no more of it.

The mole ratio between `NaI` and `NaCl` is `2:2=1`, so if `1.67*10^-2 \ "mol of" \ NaI` are used, then also `1.67*10^-2 \ "mol of" \ NaCl` are created.

Now, the question asks to find us the mass produced, so we need to find the amount in grams.

We use the equation:

`"mass"="moles"*"molar mass"`

The molar mass of `NaCl` is `58.4 \ "g/mol"`.

And so,

`m_(NaCl)=1.67*10^-2color(red)cancelcolor(black)"mol"*(58.4 \ "g")/(color(red)cancelcolor(black)"mol")=0.97528 \ "g"`

Now, I will round it off to one significant figure, like the values stated for the reactants.

`0.97528 \ "g"~~1 \ "g"`

Answer 2

See Below

Explanation:

So this boils down to a limiting reagent problem.
First, we are going to see how much NaCl could be made if we only consider `Cl_2`

Second, we'll see how much NaCl could be made if we only consider NaI. Whichever is the least amount possible, then that reagent is the one we ran out of first....it is limiting, and it'll tell us how much can be made.
Everything has to be done in moles.

We'll check `Cl_2`
5.0g `Cl_2 xx ("1mole Cl_2"/"70.9g")` = 0.0705 moles `Cl_2`

This will make NaCl:
0.0705 moles `Cl_2 xx ("2 mol NaCl"/"1 mol Cl_2")`
= 0.1410 moles NaCl

Now we'll check NaI
`5g NaI xx ("1mole NaI"/"149.89g")` = 0.0333 moles NaI
This will make NaCl:
0.0333 moles `NaI xx ("2 mol NaCl"/"2 mol NaI")`
NaCl moles = 0.0333 moles

With the `Cl_2`, we can make 0.1410 moles NaCl. Using the NaI, we can only make 0.0333 moles of NaCl. This means we will run out of NaI....NaI is our limiting reagent, and it tells us that using 5g Cl2 and 5g of NaI, we can only make 0.0333 moles NaCl.

Mass of NaCl made:
`0.0333 mol xx ("58.4g NaCl"/"1 mol NaCl")` = 1.95 g NaCl