Answer:
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 :
[tex]2Na + Cl_{2}\rightarrow 2NaCl[/tex]
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 =
[tex]\frac{70.99}{45.8}[/tex] 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 =
[tex]\frac{58.4}{22.9}[/tex] = 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
