Chemistry, 23.07.2019 08:30 anaiah13

In reference to item 13, suppose synthesis of al2o3 is done in the lab using 20.50 g of al and excess of o2. upon purification, 30.33 grams of al2o3 is produced. what is the percent yield of al2o3?

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Answer from: 61497

The answer is 78.31%
Solution:
Aluminum reacts with oxygen to form aluminum oxide as shown in the chemical equation
4Al + 3O2 → 2Al2O3

We can see that 4 moles of aluminum is required to produce 2 moles aluminum oxide. Therefore we can calculate for the mass of aluminum oxide using the molar masses of aluminum and aluminum oxide:
mass of Al2O3 = 20.50 g Al (1 mol Al / 26.982 g Al) (2 mol Al2O3 / 4 mol Al) (101.96 g Al2O3 / 1 mol Al2O3)
= 38.733 g Al2O3

If only 30.33 grams of aluminum oxide formed under the conditions the percentage yield is
percent yield = (actual mass/theoretical mass) x 100
= (30.33 / 38.733) x 100
= 78.31%

Answer from: alex12everett

The balanced equation for the above reaction is as follows;
4Al + 3O₂ --> 2Al₂O₃
stoichiometry of Al to Al₂O₃ is 4:2
the number of Al moles reacted - 20.50 g/ 27 g/mol = 0.76 mol
we have been told that O₂ was provided in excess, therefore Al is the limiting reactant.
Amount of product formed depends on amount of limiting reactant present.
number of Al₂O₃ moles formed are - 0.76 / 2 = 0.38 mol
therefore mass of Al₂O₃ formed is - 0.38 mol x 102 g/mol = 38.76 g
theoretical yield is 38.76 g
but actual yield is 30.33 g
the percentage yield = actual yield / theoretical yield x 100 %
percentage yield = 30.33 g/ 38.76 x 100% = 78.3 %
percentage yield = 78.3 %

Answer from: Quest

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

Answer from: Quest

Ais the correct answer

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In reference to item 13, suppose synthesis of al2o3 is done in the lab using 20.50 g of al and exces...

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