Urgent: in the reaction below, what is the maximum mass of nacl that can be formed when 28 grams of sodium reacts with 55 grams of chlorine?

2na+cl_2 -> 2nacl

1)45.3 g nacl
2)58.45 g nacl
3)71.2 g nacl
4)90.7 g na cl

Maximum mass of NaCl that can be formed =

3) 71 .2 g NaCl

Explanation:

Limiting reagent : The substance that is consumed in the reaction . The limiting reagent predicts the amount of product produced.

Molar mass = mass of substance present in 1 mole of molecule.

mass of Cl = 35.43 u

Molar Mass of Cl2 = 2 x 35.43 = 70.99 g

1 mole Cl2 = 70 .99 g

Molar Mass of Na = 22.99 g

1 mole of Na = 22.99 g

Molar Mass of NaCl =  58.54 g

1 mole of NaCl = 58 .54 g

The balanced equation is :

First, find the limiting reagent in this equation:

2 mole of Na need = 1 mole of Cl2

2 x 22.9 gram Na need = 70.99 gram of Cl2

45.8 gram Na need = 70.99 gram of Cl2

1 gram  Na need =

gram Cl2

1 gram  Na need = 1.55 gram Cl2

28 gram Na needs = 1.55 x 28 gram Cl2

28 gram Na needs = 43.4 gram Cl2

Cl2 in the reaction = 55 gram

Cl2 Needed = 43.5 gram

So , Cl2 is excess and Na is the limiting reagent (it is less than needed)

Now , Calculate NaCl produced from Na only

(Cl2 is of no use because it is excess reagent , it can't predict the amount of NaCl)

2 mole of Na will form = 2 moles of NaCl

So, 1 mole of Na = 1 mole of NaCl (use Molar mass of Na and NaCl)

22.9 gram of Na will form  = 58.4 gram NaCl

1  gram of Na =

= 2.55

1  gram of Na will produce = 2.55 gram of NaCl

28 gram of Na = 28 x 2.55 gram of NaCl

= 71.40 gram of NaCl

28 gram of Na = 71.40 gram of NaCl

what is the question?

c. electrical charges of ionic bonds.