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Chapter 2: Problem 4

A 1.446 g sample of potassium reacts with 8.178 g of chlorine to produce potassium chloride as the only product. After the reaction, 6.867 g of chlorine remains unreacted. What mass of potassium chloride was formed?

Short Answer

Expert verified
The mass of potassium chloride formed is 2.757 g.

Step by step solution

01

Calculate the total weight of potassium and chlorine before the reaction

Add the weight of potassium and chlorine to find the total weight before the reaction:\(1.446 g (potassium) + 8.178 g (chlorine) = 9.624 g\)
02

Find the weight of chlorine that reacted

Subtract the weight of chlorine that remained unreacted from the total weight of chlorine before the reaction to find the weight of chlorine that reacted: \(8.178 g (chlorine initially) - 6.867 g (chlorine unreacted) = 1.311 g (chlorine that reacted)\)
03

Calculate the weight of produced potassium chloride

Add the weight of the reacted potassium and chlorine to find the weight of potassium chloride that was formed: \(1.446 g (potassium) + 1.311 g (chlorine) = 2.757 g\)

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Key Concepts

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

Chemical Reactions
Chemical reactions involve transforming the starting substances, known as reactants, into new substances called products. In the exercise, potassium ( K ) reacts with chlorine ( Cl_2 ) to produce potassium chloride ( KCl ). During a chemical reaction:
  • Atoms are rearranged to form new products.
  • No new atoms are created or destroyed.
  • The properties of the products can be vastly different from those of the reactants.
In our context, potassium and chlorine combine in a specific way to generate potassium chloride, which has unique properties compared to its individual components.
Mass Conservation
The principle of mass conservation states that the mass of the reactants in a chemical reaction will equal the mass of the products. This means the total mass remains unchanged throughout the reaction. In the example:
  • Total initial mass = Mass of potassium + Mass of chlorine = 1.446 g + 8.178 g = 9.624 g
  • Total mass of products and unreacted substances = Mass of KCl + Mass of unreacted chlorine = 2.757 g + 6.867 g = 9.624 g
This confirms that mass is conserved, ensuring the calculations align with this fundamental chemical law. Understanding mass conservation helps in balancing chemical equations and predicting the amounts of products formed.
Potassium Chloride
Potassium chloride ( KCl ) is the compound produced in the exercise. This substance forms when potassium and chlorine react:
  • It is an ionic compound, consisting primarily of potassium ions ( K^+ ) and chloride ions ( Cl^- ).
  • It is commonly used in agriculture and medicine.
Potassium chloride has a crystalline structure and dissolves easily in water. When solving for the amount of KCl formed, understanding chemical formulas and stoichiometry is crucial, as these guide how reactants combine to form products.

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Most popular questions from this chapter

In Example \(2-1,\) we established that the mass ratio of magnesium to magnesium oxide is 0.455 g magnesium/ \(0.755 \mathrm{g}\) magnesium oxide. (a) What is the ratio of oxygen to magnesium oxide, by mass? (b) What is the mass ratio of oxygen to magnesium in magnesium oxide? (c) What is the percent by mass of magnesium in magnesium oxide?

Samples of pure carbon weighing \(3.62,5.91,\) and \(7.07 \mathrm{g}\) were burned in an excess of air. The masses of carbon dioxide obtained (the sole product in each case) were \(13.26,21.66,\) and \(25.91 \mathrm{g},\) respectively. (a) Do these data establish that carbon dioxide has a fixed composition? (b) What is the composition of carbon dioxide, expressed in \% C and \% O, by mass?

Determine the only possible isotope (E) for which the following conditions are met: \(\bullet\)The mass number of \(\mathrm{E}\) is 2.50 times its atomic number. \(\bullet\)The atomic number of \(\mathrm{E}\) is equal to the mass number of another isotope (Y). In turn, isotope Y has a neutron number that is 1.33 times the atomic number of \(Y\) and equal to the neutron number of selenium- 82 .

Determine (a) the number of \(\mathrm{Kr}\) atoms in a 5.25 -mg sample of krypton (b) the molar mass, \(M,\) and identity of an element if the mass of a \(2.80 \times 10^{22}\) -atom sample of the element is \(2.09 \mathrm{g}\) (c) the mass of a sample of phosphorus that contains the same number of atoms as \(44.75 \mathrm{g}\) of magnesium

Within the limits of experimental error, show that the law of conservation of mass was obeyed in the following experiment: \(10.00 \mathrm{g}\) calcium carbonate (found in limestone) was dissolved in 100.0 mL hydrochloric acid \((d=1.148 \mathrm{g} / \mathrm{mL}) .\) The products were \(120.40 \mathrm{g}\) solution (a mixture of hydrochloric acid and calcium chloride) and 2.22 L carbon dioxide gas \((d=1.9769 \mathrm{g} / \mathrm{L})\)
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