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Alexandra [31]
3 years ago
14

A student performs a reaction that makes aluminum oxide. According to her calculations, she should expect to make 115.2 grams. S

he actually produces 66.9 grams. What is her percent yield?
Chemistry
1 answer:
tiny-mole [99]3 years ago
7 0

Answer:

The percentage of the student is 58.17%.

Explanation:

Expected yield of aluminum oxide = 115.2 g

Actual yield of aluminum oxide produced =66.9 g

The percentage yield is calculated by dividing actual yield by expected yield and then multiplying it with hundred.

Percentage yield:

\frac{\text{Actual yield}}{\text{Expected yield}}\times 100

\% Yield=\frac{66.9 g}{115.2 g}\times 100=58.17\%

The percentage of the student is 58.17%.

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When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.
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Answer:

Mass of CaCl₂ =  20 g

CaCO is presewnt in excess.

Mass of of CaCO₃ remain unreacted =  7.007 g

Explanation:

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

Chemical equation:

CaCO₃ + 2HCl  →  CaCl₂  + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

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Number of moles of CaCO₃ = 0.25 mol

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Number of moles of  HCl = Mass /molar mass

Number of moles of HCl = 13.0 g / 36.5 g/mol

Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

                  CaCO₃         :               CaCl₂

                    1                 :               1

                 0.25              :            0.25

                HCl                :                CaCl₂

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ =  20 g

The calcium carbonate is present in excess.

                HCl                :                CaCO₃

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

So, 0.18 moles react with 0.36 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 -0.18

The moles of CaCO₃ remain unreacted = 0.07 mol

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.07 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted =  7.007 g

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