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o-na [289]
3 years ago
15

Determine the LIMITING reactant in the following balanced equation:

Chemistry
1 answer:
padilas [110]3 years ago
5 0

Answer:

KBr is limiting reactant.

Explanation:

Given data:

Mass of  KBr =4g

Mass of Cl₂ = 6 g

Limiting reactant = ?

Solution:

Chemical equation:

2KBr + Cl₂      →    2KCl + Br₂

Number of moles of KBr:

Number of moles = mass/molar mass

Number of moles = 4 g/ 119 gmol

Number of moles = 0.03 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 6 g/ 70 gmol

Number of moles = 0.09 mol

Now we will compare the moles of reactant with product.

              KBr            :            KCl

                2              :              2

            0.03            :            0.03

             KBr            :              Br₂

                2             :               1

             0.03           :          1/2×0.03= 0.015

               Cl₂             :            KCl

                 1              :              2

            0.09            :           2/1×0.09 = 0.18

               Cl₂             :              Br₂

                1              :               1

             0.09           :            0.09

Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂  is present in excess.

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The reaction between NaOH and Cu(NO₃)₂ is as follows

2NaOH + Cu(NO₃)₂ ---> 2NaNO₃ + Cu(OH)₂


Q1) stoichiometry of NaOH to Cu(NO₃)₂ is 2:1

this means that 2 mol of NaOH reacts with 1 mol of Cu(NO₃)₂

the mass of Cu(NO₃)₂ reacted - 0.8024 g 

molar mass of Cu(NO₃)₂ is 187.56 g/mol

therefore the number of Cu(NO₃)₂ moles that have reacted

 - 0.8024 g/ 187.56 g/mol = 0.00427 mol

according to the stoichiometry , number of NaOH moles - 0.00427 mol x 2 

then number of NaOH moles that have reacted - 0.00855 mol

In a 3.0 M NaOH solution, 3 moles are in 1000 mL of solution

Then volume required for 0.00855 mol - 1000 x 0.00855 /3 = 2.85 mL

2.85 mL of 3.0 M NaOH is required for this reaction


Q2) Assuming that there's 100 % yield of Cu(OH)₂ , we can directly calculate the mass of Cu(OH)₂ formed from the number of moles of reactants that were used up. 

Stoichiometry of Cu(NO₃)₂ to Cu(OH)₂ is 1:1

this means that 1 mole of Cu(NO₃)₂ gives a yield of  1 mole of Cu(OH)₂

the number of Cu(NO₃)₂ moles that reacted - 0.00427 mole 

Therefore an equal amount of moles of Cu(OH)₂ were formed

Then amount of Cu(OH)₂ moles produced - 0.00427 mol

Mass of Cu(OH)₂ formed - 0.00427 mol x 97.56 g/mol = 0.42 g 

A mass of 0.42 g of Cu(OH)₂ was formed in this reaction.

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