If the yield of chloroform obtainable from acetone and bleaching powder is \(75 \%\), what mass of acetone is required for producing \(30 \mathrm{~g}\) of chloroform? (a) \(40 \mathrm{~g}\) (b) \(9.4 \mathrm{~g}\) (c) \(10.92 \mathrm{~g}\) (d) \(14.56 \mathrm{~g}\)

Short Answer

Expert verified
40 g

Step by step solution

01

Understand the Problem

We need to determine the mass of acetone needed to produce 30 g of chloroform with a 75% yield. The yield percentage tells us that only 75% of the theoretical amount of acetone will be converted into chloroform.
02

Calculate Theoretical Yield

First, calculate the theoretical yield—the amount of chloroform that would be produced if the reaction were 100% efficient. To do this, use the percentage yield formula: Actual yield = (Percentage Yield / 100) * Theoretical yield. Rearrange this formula to find the theoretical yield: Theoretical yield = Actual yield / (Percentage Yield / 100).
03

Calculate Mass of Acetone Needed for Theoretical Yield

Using the stoichiometry of the reaction, which is not given in the problem, we would normally determine the ratio of chloroform to acetone to calculate the mass of acetone needed theoretically. However, since we're only given percentages and actual yield, we can infer that for a 100% yield, the mass of acetone is equal to the mass of chloroform produced. The calculated theoretical yield of chloroform is thus equal to the required mass of acetone for a 100% yield.
04

Calculate Mass of Acetone Needed for Actual Yield

Now, calculate the mass of acetone needed to obtain the actual yield (30 g of chloroform) using the 75% yield. Divide the theoretical yield of chloroform by 0.75 (which is 75% expressed as a decimal) to find the mass of acetone required: Mass of Acetone = Theoretical yield / 0.75.
05

Perform the Calculation

Substitute the actual yield (30 g of chloroform) into the rearranged percentage yield formula to find the theoretical yield: Theoretical yield = 30 g / 0.75 = 40 g. So, the mass of acetone required is 40 g.

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Key Concepts

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

Theoretical Yield Calculation
Understanding the theoretical yield is an essential part of stoichiometry, a fundamental concept in chemistry that deals with the quantitative aspects of chemical reactions. Essentially, the theoretical yield is the maximum amount of product that can be generated from a given amount of reactants in a chemical reaction, assuming that the reaction goes to completion without any losses.

When calculating the theoretical yield, we start by looking at the balanced chemical equation to understand the molar ratio between the reactants and the products. If the balanced equation is not provided, as can be the case in certain textbook exercises, assumptions might have to be made based on the information available. In the provided exercise, an assumption is made that the mass of acetone, if converted with 100% efficiency, would equal the mass of chloroform produced.

This equality simplifies the calculation, as we can directly use the mass of the desired product to establish what the theoretical yield would be if the conversion was fully efficient. From there, we can use this theoretical yield as a reference point to determine how much of the reactant is needed in a real-world scenario where the yield is less than 100%, demonstrating the practical application of theoretical yield calculations in laboratory and industrial settings.
Percentage Yield Formula
Percentage yield is a measure of the efficiency of a chemical reaction and is calculated by comparing the actual yield (the amount of product actually obtained) to the theoretical yield (the amount that would be obtained if the reaction were 100% efficient). The percentage yield formula is a simple, yet powerful tool for chemists to evaluate how well a reaction has proceeded.

The formula is given as follows:
\[\text{Percentage Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\]
To apply this formula, chemists first determine the actual yield, usually by weighing the product, and then calculate the theoretical yield as we did in the step-by-step solution. The exercise provided a unique opportunity to flip this formula. Instead of starting with the actual yield and the percentage to find the theoretical yield, it rearranged the formula to determine the mass of acetone needed when the percentage yield was given, demonstrating the versatility of the percentage yield formula in various stoichiometry problems.
Chemical Reaction Stoichiometry
Chemical reaction stoichiometry is the cornerstone of the science that quantifies these relationships in chemical reactions. It involves the use of a balanced chemical equation to determine the proportions of reactants and products involved in a chemical reaction.

Understanding stoichiometry is crucial for predicting how much of each reactant is needed and what quantity of product will be formed under certain conditions. In practical terms, it helps chemists design reactions, to be efficient and cost-effective by minimizing waste and optimizing the use of resources.

In the exercise, the chemical reaction wasn't fully described, so stoichiometry was inferred from the given data. Typically, one would identify the molar ratios from the balanced chemical equation and apply it to the mass of reactants and products. But the prompt provided a simplified scenario where 1 gram of acetone yields 1 gram of chloroform at 100% efficiency. This assumption allowed the calculation to progress without the actual stoichiometric details, demonstrating the necessity of a clear understanding of stoichiometry for all levels of chemistry.

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Most popular questions from this chapter

The empirical formula of an organic gaseous compound containing carbon and hydrogen is \(\mathrm{CH}_{2}\). The volume occupied by certain mass of this gas is exactly half of the volume occupied by the same mass of nitrogen gas under identical conditions. The molecular formula of the organic gas is (a) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (b) \(\mathrm{CH}_{2}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{12}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{8}\)

Equal masses of oxygen, hydrogen and methane are taken in identical conditions. What is the ratio of the volumes of the gases under identical conditions? (a) \(16: 1: 8\) (b) \(1: 16: 2\) (c) \(1: 16: 8\) (d) \(2: 16: 1\)

To determine soluble (free) \(\mathrm{SiO}_{2}\) in a rock. an alkaline extraction was carried out, as a result of which there was found \(1.52 \%\) of \(\mathrm{SiO}_{2}\) in the extract and also \(1.02 \%\) of \(\mathrm{Al}_{2} \mathrm{O}_{3} .\) Considering that, apart from the free \(\mathrm{SiO}_{2}\), the extract also contained the \(\mathrm{SiO}_{2}\) that had passed into it from Kaolin \(\left(2 \mathrm{SiO}_{2} \cdot \mathrm{Al}_{2} \mathrm{O}_{3}\right)\), the percentage of free \(\mathrm{SiO}_{2}\) in the rock being analysed is \((\mathrm{Si}=28\), \(\mathrm{Al}=27\) ) (a) \(1.20\) (b) \(0.32\) (c) \(0.50\) (d) \(1.52\)

An amount of \(1.0 \times 10^{-3}\) moles of \(\mathrm{Ag}^{+}\) and \(1.0 \times 10^{-3}\) moles of \(\mathrm{CrO}_{4}^{2-}\) reacts together to form solid \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\). What is the amount of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) formed? \((\mathrm{Ag}=108, \mathrm{Cr}=52)\) (a) \(0.332 \mathrm{~g}\) (b) \(0.166 \mathrm{~g}\) (c) \(332 \mathrm{~g}\) (d) \(166 \mathrm{~g}\)

The fractional abundance of \(\mathrm{Cl}^{35}\) in a sample of chlorine containing only \(\mathrm{Cl}^{35}\) (atomic weight \(=34.9\) ) and \(\mathrm{Cl}^{37}\) (atomic weight \(=36.9\) ) isotopes, is \(0.6\). The average mass number of chlorine is (a) \(35.7\) (b) \(35.8\) (c) \(18.8\) (d) \(35.77\)

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