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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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Mr = 27.2
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Repeat units = 27.2 / 14 ≈ 2

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Combustion equation:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

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Answer:

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A mixture of hydrochloric and sulfuric acids is prepared so that it contains 0.315 M HCl and 0.125 M H2SO4. What volume of 0.55
Xelga [282]

<u>Answer:</u> The volume of NaOH required is 402.9 mL

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

  • <u>For HCl:</u>

Molarity of HCl solution = 0.315 M

Volume of solution = 503.4 mL = 0.5034 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

0.315M=\frac{\text{Moles of HCl}}{0.5034L}\\\\\text{Moles of HCl}=(0.315mol/L\times 0.5034L)=0.1586mol

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Volume of solution = 503.4 mL = 0.5034 L

Putting values in equation 1, we get:

0.125M=\frac{\text{Moles of }H_2SO_4}{0.5034L}\\\\\text{Moles of }H_2SO_4=(0.125mol/L\times 0.5034L)=0.0630mol

As, all of the acid is neutralized, so moles of NaOH = [0.1586 + 0.0630] moles = 0.2216 moles

Molarity of NaOH solution = 0.55 M

Moles of NaOH = 0.2216 moles

Putting values in equation 1, we get:

0.55M=\frac{0.2216}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.2216}{0.55}=0.4029L=402.9mL

Hence, the volume of NaOH required is 402.9 mL

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