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2 years ago

if 14.0 g of aluminium reacts with excess sulfuric acid to produce 75.26 g of aluminium sulfate, what is the percent yield?

Chemistry

1 answer:

vlada-n [284]2 years ago

7 0

Taking into account definition of percent yield, the percent yield for the reaction is 84.88%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Al + 3 H₂SO₄ → Al₂(SO₄)₃ + 3 H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Al: 2 moles
H₂SO₄: 3 moles
Al₂(SO₄)₃. 1 mole
H₂: 3 moles

The molar mass of the compounds is:

Al: 27 g/mole
H₂SO₄: 98 g/mole
Al₂(SO₄)₃: 342 g/mole
H₂: 2 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Al: 2 moles ×27 g/mole= 54 grams
H₂SO₄: 3 moles ×98 g/mole= 294 grams
Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
H₂: 3 moles ×2 g/mole= 6 grams

<h3>Mass of aluminium sulfate formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 54 grams of aluminium form 342 grams of aluminium sulfate, 14 grams of aluminium form how much mass of aluminium sulfate?

$mass of aluminium sulfate=\frac{14 grams of aluminium x342 grams of aluminium sulfate}{54 grams of aluminium}$

<u><em>mass of aluminium sulfate= 88.67 grams</em></u>

Then, 88.67 grams of aluminium sulfate can be produced if 14.0 g of aluminium reacts with excess sulfuric acid.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

$percent yield=\frac{actual yield}{theorical yield} x100$

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Percent yield for the reaction in this case</h3>

In this case, you know:

actual yield= 75.26 grams
theorical yield= 88.67 grams

Replacing in the definition of percent yields:

$percent yield=\frac{75.26 grams}{88.67 grams} x100$

Solving:

<u><em>percent yield= 84.88%</em></u>

Finally, the percent yield for the reaction is 84.88%.

Learn more about

the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

percent yield:

brainly.com/question/14408642

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Explanation:

Given data:

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Formula:

Percent error = theoretical value- actual value/ actual value × 100

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In the attached picture you have the answer for this.

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3 years ago

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Answer:

It would change the charge of the atom.

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3 years ago

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3 years ago

2 NaOH (s) + CO2(g) → Na2CO3 (s) + H20 (I)

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<h3>Answer:</h3>

16.7 g H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)

[Given] 1.85 mol NaOH

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol NaOH → 1 mol H₂O

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 1.85 \ mol \ NaOH(\frac{1 \ mol \ H_2O}{2 \ mol \ NaOH})(\frac{18.02 \ g \ H_2O}{1 \ mol \ H_2O})$
Multiply/Divide: $\displaystyle 16.6685 \ g \ H_2O$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

16.6685 g H₂O ≈ 16.7 g H₂O

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