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Chapter 9: Problem 61

According to his prelaboratory theoretical yield calculations, a student's experiment should have produced \(1.44 \mathrm{~g}\) of magnesium oxide. When he weighed his product after reaction, only \(1.23 \mathrm{~g}\) of magnesium oxide was present. What is the student's percent yicld?

Short Answer

Expert verified
The student's percent yield of magnesium oxide in the experiment is 85.42%. This is calculated using the formula: Percent Yield = (Actual Yield / Theoretical Yield) 脳 100%, where Actual Yield is 1.23 g and Theoretical Yield is 1.44 g.

Step by step solution

01

Identify the given values

We are given the following values: - Theoretical Yield of magnesium oxide: 1.44 g - Actual Yield of magnesium oxide: 1.23 g
02

Apply the percent yield formula

Now, we will use the given values and plug them into the percent yield formula: Percent Yield = (Actual Yield / Theoretical Yield) 脳 100%
03

Calculate the percent yield

Substitute the given values and compute the result: Percent Yield = (1.23 g / 1.44 g) 脳 100%
04

Simplify the expression and find the answer

Calculate the division and multiply the result by 100%: Percent Yield = (0.8542) 脳 100% Percent Yield = 85.42% So, the student's percent yield of magnesium oxide in the experiment is 85.42%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Theoretical Yield
Theoretical yield is an important concept in chemistry. It refers to the maximum amount of product that can be produced in a chemical reaction. This prediction is based on the balanced equation and the initial quantities of reactants used. The theoretical yield is calculated by assuming that:
  • The reaction goes to completion.
  • All reactants are converted into the desired product without any side reactions.
  • No losses occur during the process.
In the context of the magnesium oxide experiment, the theoretical yield was calculated to be 1.44 grams. This means that, theoretically, if every atom of magnesium reacted with oxygen perfectly, the student should have obtained exactly 1.44 grams of magnesium oxide. However, achieving the theoretical yield in real-world lab settings is extremely rare due to various unavoidable factors.
Actual Yield
Actual yield refers to the amount of product that is actually produced when the chemical reaction is carried out. It often varies from the theoretical yield due to a range of factors, including:
  • Incomplete reactions where not all reactants form the desired product.
  • Side reactions that can produce different products.
  • Loss of product during processes like filtration or transfer.
In the scenario of the magnesium oxide experiment, the actual yield was 1.23 grams. This value represents what the student physically obtained after the reaction. Differences between actual and theoretical yields are typical in experiments and highlight the practical challenges in converting reactants to products efficiently.
Magnesium Oxide
Magnesium oxide (MgO) is a compound formed by the combination of magnesium and oxygen. It is commonly produced through a simple reaction where magnesium burns in the presence of oxygen:
\[ 2 \, Mg + O_{2} \rightarrow 2 \, MgO \]Magnesium oxide is known for its white, powdery form and high melting point. It is widely used across various industries:
  • As a refractory material in high-temperature furnaces.
  • In agriculture as a magnesium source for animals and plants.
  • In medicine, used to treat indigestion and as a magnesium supplement.
The study of this compound in experiments helps students understand chemical reactions and concepts like yield calculations, due to its straightforward reaction properties and easily measurable results.

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Most popular questions from this chapter

Calcium carbide, \(\mathrm{CaC}_{2},\) can be produced in an electric furnace by strongly heating calcium oxide (lime) with carbon. The unbalanced cquation is $$ \mathrm{CaO}(s)+\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}(g) $$ Calcium carbide is useful because it reacts readily with water to form the flammable gas acctylene, \(\mathrm{C}_{2} \mathrm{H}_{2},\) which is used extensively in the welding industry. The unbalanced equation is $$ \mathrm{CaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s) $$ What mass of acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2},\) would be produced by complete reaction of \(3.75 \mathrm{~g}\) of calcium carbide?

Consider the balanced chemical equation $$ 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ What mole ratio would you use to calculate how many moles of oxygen gas would be needed to react completely with a given number of moles of aluminum metal? What mole ratio would you use to calculate the number of moles of product that would be expected if a given number of moles of aluminum metal reacts completely?

For each of the following unbalanced chemical equations, suppose that exactly \(1.00 \mathrm{~g}\) of each reactant is taken. Determine which reactant is limiting, and calculate what mass of the product in boldface is expected (assuming that the limiting reactant is completely consumed). a. \(\mathrm{CS}_{2}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{SO}_{2}(g)\) b. \(\mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CN}_{2} \mathrm{H}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{MnO}_{2}(s) \rightarrow \mathrm{MnO}(s)+\mathbf{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{I}_{2}(l)+\mathrm{Cl}_{2}(g) \rightarrow \mathbf{I C l}(g)\)

For the balanced chemical equation for the combination reaction of hydrogen gas and oxygen gas $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ explain why we know that \(2 \mathrm{~g}\) of \(\mathrm{H}_{2}\) reacting with \(1 \mathrm{~g}\) of \(\mathrm{O}_{2}\) will not result in the production of \(2 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\)

Using the average atomic masses given inside the front cover of the text, calculate the mass in grams of cach of the following samples. a. 5.0 moles of nitric acid b. 0.000305 mole of mercury c. \(2.31 \times 10^{-5}\) mole of potassium chromate d. 10.5 moles of aluminum chloride e. \(4.9 \times 10^{4}\) moles of sulfur hexafluoride f. 125 moles of ammonia g. 0.01205 mole of sodium peroxide
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