/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Ammonia gas can be prepared by t... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 20

Ammonia gas can be prepared by the following reaction: \(\mathrm{CaO}(\mathrm{s})+2 \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) \rightarrow\) $$ 2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CaCl}_{2}(\mathrm{s}) $$ If \(112 \mathrm{g}\) of \(\mathrm{CaO}\) and \(224 \mathrm{g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) are mixed, the theoretical yield of \(\mathrm{NH}_{3}\) is \(68.0 \mathrm{g}\) (Study Question 12 ). If only \(16.3 \mathrm{g}\) of \(\mathrm{NH}_{3}\) is actually obtained, what is its percent yield?

Short Answer

Expert verified
The percent yield of \(\text{NH}_3\) is approximately 24\%.

Step by step solution

01

Identify Known Quantities

We are given the theoretical yield of ammonia as \(68.0 \, \text{g}\) and the actual yield as \(16.3 \, \text{g}\). To find the percent yield, we'll use these values.
02

Use the Percent Yield Formula

The formula for percent yield is \(\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\). Insert the known values into the formula.
03

Calculate the Percent Yield

Substitute the actual yield of \(16.3 \, \text{g}\) and the theoretical yield of \(68.0 \, \text{g}\) into the formula: \[\text{Percent Yield} = \left(\frac{16.3}{68.0}\right) \times 100\%\]Perform the division to find the proportion.
04

Compute the Result

Calculate \(\frac{16.3}{68.0} = 0.2397\). Multiply by 100 to convert to a percentage: \[0.2397 \times 100 = 23.97\%\].

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Key Concepts

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

Theoretical Yield
Theoretical yield represents the maximum amount of product that can be generated from a chemical reaction, based on the complete conversion of the limiting reactant. This is a calculated expectation, determined using the balanced chemical equation and stoichiometry.
Calculating the theoretical yield involves several steps:
  • Identify the balanced chemical equation. Here, we use \( \mathrm{CaO} + 2\ \mathrm{NH}_{4} \mathrm{Cl} \rightarrow 2\ \mathrm{NH}_{3} + \mathrm{H}_{2} \mathrm{O} + \mathrm{CaCl}_{2} \).
  • Determine the limiting reactant. This is the reactant that will run out first, limiting the amount of product formed.
  • Use stoichiometry to calculate the amount of product formed from the limiting reactant. Apply molar ratios from the balanced equation to determine the theoretical yield.
Understanding the theoretical yield is key to evaluating how efficient a reaction is likely to be under ideal conditions.
Actual Yield
Actual yield refers to the real amount of product obtained from a reaction. This is often less than the theoretical yield due to various factors such as incomplete reactions, impurities, or measurement errors.
Achieving the actual yield involves:
  • Conducting the chemical reaction as accurately as possible.
  • Measuring the product obtained carefully to ensure precision.
Several reasons might lead the actual yield to differ from the theoretical yield:
  • Side reactions that consume some of the reactants.
  • Loss of product during collection or transfer.
  • Unexpected environmental factors affecting reaction dynamics.
Understanding the actual yield helps chemists and students acknowledge real-world limitations in experiments.
Chemical Reaction Stoichiometry
Chemical reaction stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It provides the coefficients that dictate the proportions of reactants needed to form products.
To utilize stoichiometry:
  • Examine the balanced chemical equation.
  • Use the molar ratios to calculate how much of each reactant is required or how much product will be generated.
  • Apply these ratios to convert between masses, moles, and volumes (if dealing with gases).
For example, in our reaction, stoichiometry tells us that 2 moles of \( \mathrm{NH}_{4} \mathrm{Cl} \) produce 2 moles of \( \mathrm{NH}_{3} \). By understanding these ratios, it鈥檚 easier to solve problems related to the yield and efficiency of reactions. Accurate stoichiometric calculations are essential for predicting yield and scaling reactions in industrial applications.

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Most popular questions from this chapter

If 38.55 mL of \(\mathrm{HCl}\) is required to titrate \(2.150 \mathrm{g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) according to the following equation, what is the concentration \((\mathrm{mol} / \mathrm{L})\) of the HCl solution? \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow\) \(2 \mathrm{NaCl}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)\)

A Phosphate in urine can be determined by spectrophotometry. After removing protein from the sample, it is treated with a molybdenum compound to give, ultimately, a deep blue polymolybdate. The absorbance of the blue polymolybdate can be measured at \(650 \mathrm{nm}\) and is directly related to the urine phosphate concentration. A 24 -hour urine sample was collected from a patient; the volume of urine was 1122 mL. The phosphate in a \(1.00 \mathrm{mL}\), portion of the urine sample was converted to the blue polymolybdate and diluted to \(50.00 \mathrm{mL} .\) A calibration curve was prepared using phosphate-containing solutions. (Concentrations are reported in grams of phosphorus (P) per liter of solution.) $$\begin{array}{lc} \text { Solution (mass P/L) } & \begin{array}{c} \text { Absorbance at } 650 \mathrm{nm} \\ \text { in a } 1.0-\mathrm{cm} \text { cell } \end{array} \\ \hline 1.00 \times 10^{-6} \mathrm{g} & 0.230 \\ 2.00 \times 10^{-6} \mathrm{g} & 0.436 \\ 3.00 \times 10^{-6} \mathrm{g} & 0.638 \\ 4.00 \times 10^{-6} \mathrm{g} & 0.848 \\ \text { Urine sample } & 0.518 \\ \hline \end{array}$$ (a) What are the slope and intercept of the calibration curve? (b) What is the mass of phosphorus per liter of urine? (c) What mass of phosphate did the patient excrete in the one-day period?

What mass of lime, CaO, can be obtained by heating \(125 \mathrm{kg}\) of limestone that is \(95.0 \%\) by mass \(\mathrm{CaCO}_{3} ?\) $$ \mathrm{CaCO}_{3}(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$

You wish to determine the weight percent of copper in a copper-containing alloy. After dissolving a \(0.251-\mathrm{g}\) sample of the alloy in acid, an excess of KI is added, and the \(\mathrm{Cu}^{2+}\) and \(1^{-}\) ions undergo the reaction $$ 2 \mathrm{Cu}^{2+}(\mathrm{aq})+5 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{CuI}(\mathrm{s})+\mathrm{I}_{3}^{-}(\mathrm{aq}) $$ The liberated \(I_{3}^{-}\) is titrated with sodium thiosulfate according to the equation \(\mathrm{I}_{3}^{-}(\mathrm{aq})+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq})\) (a) Designate the oxidizing and reducing agents in the two reactions above. (b) If 26.32 mL of \(0.101 M N a_{2} S_{2} O_{3}\) is required for titration to the equivalence point, what is the weight percent of Cu in the alloy?

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