The chemical reaction would be as follows:
<span>2Na + S → Na2S
We are given the amount of the reactants to be used in the reaction. We use these to calculate the amount of product. We do as follows:
45.3 g Na ( 1 mol / 22.99 g ) = 1.97 mol Na
105 g S ( 1 mol / 32.06 g ) = 3.28 mol S
The limiting reactant would be Na. We calculate as follows:
1.97 mol Na ( 1 mol Na2S / 2 mol Na ) (78.04 g / mol ) = 76.87 g Na2S produced</span>
When comparing single bonds between atoms of comparable types, the stronger the bond is, the bigger the atom, the weaker it is.
The length of the X-H bond lengthens while the strength of the bond shortens with increasing halogen size (F-H strongest, I-H weakest). When comparing single bonds between atoms of similar sorts, the larger the atom, the weaker the bond. It can be explained by the fact that less energy is required to break the bond the bigger the atom's atomic size. The force of attraction from the nucleus to the outermost orbit will be less for iodine since it has a larger atom than the other elements in the group.
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We know that to relate solutions of with the factors of molarity and volume, we can use the equation:

**
NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.
So now we can assign values to these variables. Let us say that the 18 M

is the left side of the equation. Then we have:

We can then solve for

:

and

or

We now know that the total amount of volume of the 4.35 M solution will be
210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.
Answer:
The answer is 2i on right hand side.
Explanation:
We should star by checking the equation from right.
First we check how many Zn r there in left hand side. Which is 1. Let us check how many Znr there in right hand side, there is 1.So Zn is balanced, and don't worry about Znplus2 on right hand side it is just the ions not how many zinc r there.
Now let us check how many I are there left hand side. Which is 2. Now how many I are there in right hand side? Only 1.
So we put 2 behind I.
Now there r 2 I on both sides.
Its simple actually.