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

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation:

the answer is d.) potassium