Question

Chlorine is used to make bleach solutions containing 5.25% NaClO (by mass). Assuming 100% yield in the reaction producing NaClO from Cl₂, how many liters of Cl₂(g) at STP will be needed to make 1000. L of bleach solution (d = 1.07 g/mL)?

Answer 1

To make 1000. L of bleach solution 1.69×10⁴ L of Cl₂(g) at STP will be needed .

What is Balanced chemical equation?

A balance chemical equation is an equation that contain same number of atoms as well as of each element on each side of reaction.

What is Density?

It is ratio of mass to the volume.

Density= $\frac{Mass}{Volume}$

Now,

   The balanced chemical equation of preparing bleach from chlorine is as follow:

Cl₂(g) + 2NaOH(aq) → NaClO(aq) + NaCl(aq) + H₂O(l)

Now,

  To calculate the Moles of NaClO in 1000L solution, first calculate it's mass.

Mass of NaClO = Density × Volume

                         =  1.07$\frac{g}{ml}$ × 1000L × $\frac{1000mL}{1L}$ ×$\frac{5.25}{100}$

                         = 56175g

Now, we have to calculate moles of NaClO

Hence,

  Moles of NaClO = $\frac{amount of NaClO}{molar mass of NaClO}$

                              =  $\frac{56175g}{74.44\frac{g}{mol} }$

                              = 754.63 mol

Now,

According to the reaction number of moles of Cl₂ is equal to the number of moles of NaClO.

Hence,

  moles of Cl₂ = 754.63 mol

Now , in order to calculate litres of Cl₂ at STP,

 We know that

 1 mol Cl₂ = 22.4 L of Cl₂

 754.63 mol of Cl₂ = 754.63 mol × 22.4 L

                               =   1.69×10⁴ L of Cl₂

Thus from the above conclusion we can say that , to make 1000. L of bleach solution 1.69×10⁴ L of Cl₂(g) at STP will be needed .

Learn more about Balanced Chemical Equation here:brainly.com/question/27135494

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