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Briefly describe (a) balancing a chemical equation; (b) preparing a solution by dilution; (c) determining the limiting reactant in a reaction.

Short Answer

Expert verified
In summary, (a) balancing a chemical equation means ensuring equal numbers of each type of atom on both sides of the equation, (b) preparing a solution by dilution implies adding solvent to a concentrated solution to reach a desired concentration, and (c) the limiting reactant of a reaction is the one that is completely consumed first.

Step by step solution

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STEP 1: Balancing a chemical equation

Balancing a chemical equation involves ensuring that there is an equal number of each type of atom on both the reactant (left-hand) and product (right-hand) sides of the equation. Typically, it is advisable to start by balancing the atoms of elements that appear in only one compound on each side of the equation, and then proceed to balance the rest.
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STEP 2: Preparing a solution by dilution

In chemistry, preparing a solution by dilution involves adding solvent to a more concentrated solution, known as a 'stock solution', to achieve a desired final concentration. The relationship between the initial and final concentrations and volumes of the solutions can be expressed as: \(C1V1 = C2V2\), where \(C1\) and \(V1\) are the concentration and volume of the original solution respectively, and \(C2\) and \(V2\) are those of the resulting diluted solution. It's important to mix the solution thoroughly after dilution to make sure the solute is evenly distributed throughout.
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STEP 3: Determining the limiting reactant

The limiting reactant in a chemical reaction is the substance that is completely consumed when the reaction is complete. To identify the limiting reactant, it's advantageous to convert all given information to moles, and then use stoichiometric ratios from the balanced equation to see which reactant produces the least amount of product. The one which 'runs out' first is the limiting reactant.

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Key Concepts

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

Balancing Chemical Equations
The first step in understanding a chemical reaction is balancing the chemical equation. To visualize this, picture a scale in equilibrium, with equal weights on both sides. Similarly, a balanced chemical equation has the same number of each type of atom on both sides of the reaction arrow. It's a fundamental principle following the Law of Conservation of Mass, indicating that matter cannot be created or destroyed in an isolated system.

When balancing equations, start by making a list of all the elements involved in the reactants and products. Then, adjust the coefficients—the numbers in front of the compounds—to ensure that the number of atoms for each element is the same on both sides. It’s like solving a puzzle with the atoms as pieces, making sure each side mirrors the other perfectly. An easy way to approach this is by first balancing the metals, nonmetals, and then hydrogen and oxygen atoms, as they are often found in multiple compounds. Remember that the coefficients represent the relative amounts of substances in the reaction and play a crucial role in the stoichiometry, which is the calculation of reactants and products in chemical reactions.
Solution Preparation by Dilution
Creating a solution of a specific concentration often requires the process of dilution. Think of dilution like adding water to juice concentrate to get the flavor just right—it's about achieving the perfect balance. In chemistry, dilution involves adding a solvent, usually water, to a concentrated stock solution. The key to successful dilution is understanding the relationship given by the equation \(C1V1 = C2V2\), where \(C1\) is the stock solution's concentration, \(V1\) is its volume, \(C2\) is the diluted solution's concentration, and \(V2\) is its volume.

By rearranging this equation, you can solve for any unknown value. It’s essential to mix the solution well to ensure uniformity throughout. If you think of each molecule of solute as a fish in a pond, after diluting, each fish needs to swim around to spread out evenly. Always add the stock solution to water rather than the reverse, for safety and to improve the dilution process. The art of dilution is a valuable skill in the laboratory, affecting the accuracy of experiments and the reproducibility of results.
Determining the Limiting Reactant
In chemical cuisine, the limiting reactant is like the ingredient you run out of first, which determines how much final product you can make. It's the substance that controls the extent of the reaction and is completely consumed when the reaction goes to completion. To find the limiting reactant, you have to enter the world of stoichiometry, a quantitative relationship between reactants and products in a balanced chemical equation.

To uncover which reactant is the limiting one, first convert all reactant quantities to moles using their molar mass. Molar mass acts like the conversion factor between grams and moles, much like a dozen relates to the number twelve. Then, apply the mole ratios derived from the balanced equation to each reactant to determine how much product they can individually produce. Imagine two chefs making sandwiches: one with plenty of bread but little cheese, the other with enough cheese but limited bread. The chef who runs out of bread first dictates the maximum number of sandwiches made—this is your limiting reactant. Identifying it is crucial for predicting the amount of product formed and for understanding reaction efficiency in the real world.

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Most popular questions from this chapter

What are the molarities of the following solutes? (a) aspartic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{5} \mathrm{NO}_{4}\right)\) if \(0.405 \mathrm{g}\) is dissolved in enough water to make \(100.0 \mathrm{mL}\) of solution (b) acetone, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O},(d=0.790 \mathrm{g} / \mathrm{mL})\) if \(35.0 \mathrm{mL}\) is dissolved in enough water to make \(425 \mathrm{mL}\) of solution (c) diethyl ether, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O},\) if \(8.8 \mathrm{mg}\) is dissolved in enough water to make 3.00 L of solution

How many moles of \(\mathrm{NO}(\mathrm{g})\) can be produced in the reaction of \(3.00 \mathrm{mol} \mathrm{NH}_{3}(\mathrm{g})\) and \(4.00 \mathrm{mol} \mathrm{O}_{2}(\mathrm{g}) ?\) $$ 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \stackrel{\Delta}{\longrightarrow} 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$

Write chemical equations to represent the following reactions. (a) Calcium phosphate is heated with silicon dioxide and carbon, producing calcium silicate \(\left(\mathrm{CaSiO}_{3}\right)\) phosphorus ( \(\mathrm{P}_{4}\) ), and carbon monoxide. The phosphorus and chlorine react to form phosphorus trichloride, and the phosphorus trichloride and water react to form phosphorous acid. (b) Copper metal reacts with gaseous oxygen, carbon dioxide, and water to form green basic copper carbonate, \(\mathrm{Cu}_{2}(\mathrm{OH})_{2} \mathrm{CO}_{3}\) (a reaction responsible for the formation of the green patina, or coating, often seen on outdoor bronze statues). (c) White phosphorus and oxygen gas react to form tetraphosphorus decoxide. The tetraphosphorus decoxide reacts with water to form an aqueous solution of phosphoric acid. (d) Calcium dihydrogen phosphate reacts with sodium hydrogen carbonate (bicarbonate), producing calcium phosphate, sodium hydrogen phosphate, carbon dioxide, and water (the principal reaction occurring when ordinary baking powder is added to cakes, bread, and biscuits).

The minerals calcite, \(\mathrm{CaCO}_{3},\) magnesite, \(\mathrm{MgCO}_{3}\) and dolomite, \(\mathrm{CaCO}_{3} \cdot \mathrm{MgCO}_{3},\) decompose when strongly heated to form the corresponding metal oxide(s) and carbon dioxide gas. A 1.000 -g sample known to be one of the three minerals was strongly heated and \(0.477 \mathrm{g} \mathrm{CO}_{2}\) was obtained. Which of the three minerals was it?

A reaction mixture contains \(1.0 \mathrm{mol} \mathrm{CaCN}_{2}\) (calcium cyanamide) and \(1.0 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\). The maximum number of moles of \(\mathrm{NH}_{3}\) produced is (a) \(3.0 ;\) (b) 2.0 (c) between 1.0 and 2.0; (d) less than 1.0. $$\mathrm{CaCN}_{2}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{CaCO}_{3}+2 \mathrm{NH}_{3}(\mathrm{g})$$

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