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shepuryov [24]
3 years ago
6

How many grams of sodium chloride decompose to yield 15 grams of chlorine gas?

Chemistry
1 answer:
defon3 years ago
8 0

Answer:

24.6g of NaCl

Explanation:

Expression of the reaction:

            2NaCl →  2Na  +  Cl₂

Given parameters:

Mass of Cl₂  = 15g

Unknown:

Mass of NaCl  = ?

Solution:

To solve this problem, we have to use mole relationships.

 Find the number of moles of the mass of the given specie;

    Number of moles  = \frac{mass}{molar mass}  

     Molar mass of Cl₂  = 2(35.5) = 71g/mol

   Number of moles  = \frac{15}{71}   = 0.21mole

Now;

 From the balanced reaction equation;

        1 mole of Cl₂ is produced from 2 moles of NaCl;

      0.21 mole of Cl₂ will be produced from 0.21 x 2 = 0.42mole of NaCl

So,

  Mass of NaCl = number of moles x molar mass

    Molar mass of NaCl  = 23 + 35.5 = 58.5g/mol

 Mass of NaCl  = 0.42 x 58.5 = 24.6g of NaCl

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Answer:

\boxed{\text{0.50 mol/L}}

Explanation:

The balanced equation is

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1. Set up an ICE table

\begin{array}{ccccc}\rm 2COF_{2} & \, \rightleftharpoons \, & \rm CO_{2} & +&\rm CF_{4}\\2.00& & 0& & 0\\-x& & +x & & +x\\2.00 - x& & x & &x \\\end{array}

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K_{c} = \dfrac{[\rm CO][ \rm CF_{4}]}{[\rm COF_{2}]^{2}} = 9.00\\\\\begin{array}{rcl}\dfrac{x^{2}}{(2.00 - x)^{2}} & = & 9.00\\\dfrac{x}{2.00 - x} & = & 3.00\\x & = &3.00(2.00 - x)\\x & = & 6.00 - 3.00x\\4.00x & = & 6.00\\x & = & \mathbf{1.50}\\\end{array}

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\text{The equilibrium concentration of COF$_{2}$ at equilibrium is $\boxed{\textbf{0.50 mol/L}}$}

Check:

\begin{array}{rcl}\dfrac{1.50^{2}}{0.50^{2}} & = & 9.00\\\\\dfrac{2.25}{0.25}& = & 9.00\\\\9.00 & = & 9.00\\\end{array}

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