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Metallic sodium reacts vigorously with liquid bromine in the following reaction:
if 1 kg of Na is brought into contact with 3 kg of liquid bromine. Presuming that the reaction is quantitative and proceeds to completion, determine the limiting reagent and the quantity of NaBr formed. Determine the amount of excess reagent remaining after the reaction is complete.

Respuesta :

cristisp93

First we look at the chemical reaction:

2 Na + Br₂ → 2 NaBr

number of moles = mass / molecular weight

mass = number of moles × molecular weight

number of moles of Na = 1000 / 23 = 43.48 moles

number of moles of Br₂ = 3000 / 160 = 18.75 moles

from the reaction:

if        2 moles of Na are reacting with 1 mole of Br₂

then   X moles of Na are reacting with 18.75 moles of Br₂

X = (18.75 × 2) / 1 = 37.5 moles of Na

The limiting reactant is bromine Br₂.

After the reaction is complete you remain with Na as excess reagent.

moles of Na = 43.48 - 37.5 = 5.98 moles

mass of Na =  5.98 × 23 = 137.54 g

if        reacting 1 mole of Br₂  produces 2 moles of NaBr

then  reacting 18.75 moles of Br₂  produces Y moles of NaBr

Y  = (18.75 × 2) / 1 = 37.5 moles of NaBr

mass of NaBr = 37.5 × 103 = 3862.5 g

quasarJose

Answer:

Explanation:

Given parameters:

Mass of Na = 1kg = 1000g

Mass of Br = 3kg = 3000g

Unkown:

Limiting reagent =?

Amount of excess reagent = ?

Quantity of NaBr formed = ?

Solution

The reaction equation can be expressed as:

                 2Na + Br₂ → 2NaBr

To find the limiting reagent

The limiting reagent is the reagent that is in short supply and it determines the amount of products that can be formed in the reaction.

To find the limiting the reagent, we establish a mole relationship between the reactants. We then check their stoichoimetric expression to see the one in excess and the limiting reagent.

Number of moles of Na = [tex]\frac{mass of Na}{molar mass of Na}[/tex]

   molar mass is expressed as g/mol and the Na should also be in grams

Number of moles = [tex]\frac{1000}{23}[/tex] = 43.48mol

Number of moles of Br₂ =  [tex]\frac{mass}{molar mass}[/tex]

    Number of moles = [tex]\frac{3000}{2x80}[/tex] = 18.75mol

From the balanced reaction equation:

     2 mole of Na combines with 1 mole of Br₂

     43.48mole of Na would require 43.48/2 mole = 21.74mole of Br₂ to react with.

But the given amount of Br we have is 18.75mole.

Therefore, the reagent in short supply is Bromine. It is the limiting reagent

Quantity of NaBr formed

To find the quantity of NaBr formed, we use the known reagent which is the limiting one to determine its number of moles and eventually the mass. The limiting reagent determines the extent of the reaction.

 From the stoichiometric equation:

          1 mole of Br₂ will produce 2 mole of NaBr

  19.23 mole Br₂ would now yield  (18.75 x2)= 37.5mole  of NaBr

Mass of NaBr = number of moles of NaBr x molar mass of NaBr

Molar mass of NaBr = 23 + 80 = 103g/mol

      Mass of NaBr = 37.5 x 103 = 3862.5g = 3.86Kg

Amount of excess reagent remaining after the complete reaction:

The amount of excess reagent would be derived from the reagent that occurs in excess which is Na:

1 mole of Br₂ reacts with 2 moles of Na

18.75mole of Br₂  will require 2x 18.75 = 37.5mole of Na

Excess mole of Na = (43.48 - 37.5)mole = 5.98mole

Mass of excess Na = number of moles of excess Na x molar mass of Na

Mass of excess Na = 5.98 x 23

Mass of excess Na = 137.54g or 0.14g

   

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