QUESTION BELOW!!?

anytime you get one of these questions in stoichiometry (like how much of this or that was produced or consumed or what is the yield or % yield), you should follow these 6 steps

.. (1) write a balanced reaction

.. (2) convert everything given to moles

.. (3) determine limiting reagent

.. (4) convert moles limiting reagent to moles product

.. (5) convert moles product to mass product.  this is theoretical yield in mass

.. (6) % yield = (actual mass recovered / theoretical mass) x 100%

the idea being, the coefficients of a balanced equation give is MOLE ratios of chemicals in the reactions.  So we can use those values to convert between moles of the different substances.

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in this case, we're given % yield and asked to find actual yield. the only difference then is in step 6.

*** 1 ***

balanced reaction

.. 1 Cl2 + 2 KBr ---> 2 KCl + 1 Br2

*** 2 ***

moles

.. 35.1L Cl2 @STP * (1 mol / 22.41L @STP) =1.566 mol Cl2

.. 1.50L KBr * (3.50 mol KBr / 1L ) = 5.25 mol KBr

*** 3 ***

limiting reagent.  There are 3 ways to do this.  In this case, the simplest way is to divide moles available by coefficient of balanced equation.  Whichever ratio is the lowest belongs to the limiting reagent

.. mol Cl2 / coeff. Cl2 = 1.566 / 1 = 1.566

.. mol KBr / coeff. KBr = 5.25 / 2 = 2.625

since 1.566 < 2.625, Cl2 is the LR

*** 4 ***

convert moles LR to moles product

.. 1.566 mol Cl2 * (1 mol Br2 / 1 mol Cl2) = 1.566 mol Br2

*** 5 ***

theoretical mass Br2

.. 1.566 mol Br2 * (159.81g Br2 / mol Br2) = 250.3g Br2

*** 6 ***

actual mass Br2

.. % yield = (actual / theoretical) * 100%

.. actual = (%yield / 100%) * theoretical

.. actual = (35% / 100%) * 250.3g = 87.60g

and finally, we want to convert that to mL

.. 87.60g Br2 * (1mL / 3.10g) = 28.2mL 

Chlorine + Potassium bromide ➜ Bromine + potassium chloride

Cl₂ + 2KBr ➜ 2KCl + Br₂

molecular weights;

2KBr = 2•119 = 238 g/mol

Cl₂ = 2•35.45 = 70.9

Br₂ = 2•79.9 = 159.8

2KCl = 2•74.55 = 149.1

check: 70.9 + 238 = 308.9

          159.8 + 149.1 = 308.9

1 mol of Cl₂ + 2 mol of KBr ➜ 2 mol of KCl + 1 mol of Br₂

70.9 g of Cl₂ + 238 g of KBr ➜ 149.1 g of KCl + 159.8 g of Br₂

35.1 L of chlorine at STP ? what is amount ?

One mole of any ideal gas at STP has a volume of 22.41L  (old def of STP, 1 atm)

35.1 L x 1 mol/22.41L = 1.566 mol

1.50 L sample of a 3.50 M solution of KBr ? what is amount ?

3.5 M is 3.5 mol/L

3.5 mol/L x 1.50 L = 5.25 mol

which is controlling amount?

1.566 mol of Cl₂ or 5.25 mol of KBr ?

mole ratio is 1:2 so 1.566 mol of Cl₂ requires 3.132 mol of KBr

so amount of Cl₂ is controlling factor.

Br₂ yield?

theoretical amount is 1:1 mol, so 1.566 mol.

35% of that is 0.548 mol

volume at STP of 0.548 mol ?

0.548 mol / 1 mol/22.41L = 0.0245 L or 24.5 mL