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Ch.4 - Chemical Quantities & Aqueous Reactions
Chapter 4, Problem 52

Many computer chips are manufactured from silicon, which occurs in nature as SiO2. When SiO2 is heated to melting, it reacts with solid carbon to form liquid silicon and carbon monoxide gas. In an industrial preparation of silicon, 155.8 kg of SiO2 reacts with 78.3 kg of carbon to produce 66.1 kg of silicon. Determine the limiting reactant and the theoretical yield.

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Write the balanced chemical equation for the reaction: \[ \text{SiO}_2 (s) + 2\text{C} (s) \rightarrow \text{Si} (l) + 2\text{CO} (g) \]
Convert the masses of \( \text{SiO}_2 \) and \( \text{C} \) to moles using their molar masses: \( \text{Molar mass of SiO}_2 = 60.08 \, \text{g/mol} \) and \( \text{Molar mass of C} = 12.01 \, \text{g/mol} \).
Calculate the moles of \( \text{SiO}_2 \) and \( \text{C} \) using the formula: \( \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \).
Determine the limiting reactant by comparing the mole ratio of \( \text{SiO}_2 \) to \( \text{C} \) from the balanced equation with the calculated moles.
Calculate the theoretical yield of silicon by using the moles of the limiting reactant and the stoichiometry of the balanced equation.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limiting Reactant

The limiting reactant is the substance that is completely consumed first in a chemical reaction, thus determining the maximum amount of product that can be formed. To identify it, one must compare the mole ratios of the reactants based on the balanced chemical equation. The reactant that produces the least amount of product is the limiting reactant.
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Theoretical Yield

The theoretical yield is the maximum amount of product that can be generated from a given amount of reactants, calculated using stoichiometry based on the balanced chemical equation. It assumes complete conversion of the limiting reactant into product without any losses. The theoretical yield is expressed in grams or moles and is essential for evaluating the efficiency of a reaction.
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Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It involves using balanced chemical equations to calculate the amounts of substances consumed and produced. Understanding stoichiometry is crucial for determining limiting reactants and theoretical yields in chemical processes.
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Related Practice
Open Question
Lead ions can be precipitated from solution with KCl according to the reaction: Pb2+ (aq) + 2 KCl(aq) → PbCl2(s) + 2 K+ (aq). When 28.5 g KCl is added to a solution containing 25.7 g Pb2+, a PbCl2 precipitate forms. The precipitate is filtered, dried, and found to have a mass of 29.4 g. Determine the percent yield for the reaction. Determine the theoretical yield of PbCl2. Determine the limiting reactant.
Textbook Question

Magnesium oxide can be made by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is: 2 Mg(s) + O2(g) → 2 MgO(s) When 10.1 g of Mg reacts with 10.5 g O2, 11.9 g MgO is collected. Determine the limiting reactant, theoretical yield, and percent yield for the reaction.

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Textbook Question

Urea (CH4N2O) is a common fertilizer that is synthesized by the reaction of ammonia (NH3) with carbon dioxide: 2 NH3(aq) + CO2(aq) → CH4N2O(aq) + H2O(l) In an industrial synthesis of urea, a chemist combines 136.4 kg of ammonia with 211.4 kg of carbon dioxide and obtains 168.4 kg of urea. Determine the limiting reactant, theoretical yield of urea, and percent yield for the reaction.

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Textbook Question

Many computer chips are manufactured from silicon, which occurs in nature as SiO2. When SiO2 is heated to melting, it reacts with solid carbon to form liquid silicon and carbon monoxide gas. In an industrial preparation of silicon, 155.8 kg of SiO2 reacts with 78.3 kg of carbon to produce 66.1 kg of silicon. Determine the percent yield for the reaction.

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Textbook Question

Calculate the molarity of each solution.

a. 3.25 mol of LiCl in 2.78 L solution

b. 28.33 g C6H12O6 in 1.28 L of solution

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Textbook Question

Calculate the molarity of each solution.

c. 32.4 mg NaCl in 122.4 mL of solution

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