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Why is it essential to use balanced chemical equations when determining the quantity of a product formed from a given quantity of a reactant?

Short Answer

Expert verified
Using balanced chemical equations is essential when determining the quantity of a product formed from a given quantity of a reactant because it ensures the Law of Conservation of Mass is obeyed and provides stoichiometric coefficients for stoichiometry calculations. These coefficients help in accurately predicting reactant amounts required and product amounts formed, allowing for efficient planning and optimization of chemical reactions while reducing waste.

Step by step solution

01

Balanced Chemical Equations

In a balanced chemical equation, the number of atoms of each element is the same on the reactant side and the product side. This ensures that the Law of Conservation of Mass is followed, meaning the total mass of reactants is equal to the total mass of products. A balanced chemical equation also represents the stoichiometric coefficients, showing the relationship between moles of reactants and products.
02

Stoichiometry

Stoichiometry is the calculation of the quantities of reactants and products in a chemical reaction. It is based on the coefficients of the balanced chemical equation, which provide the molar proportions of the substances involved. These coefficients help us in determining the mass of reactants needed to produce a desired mass of products or to calculate the mass of products formed from a given mass of reactants.
03

Importance of Balanced Chemical Equations

When determining the quantity of a product formed from a given quantity of a reactant, it's essential to use a balanced chemical equation because: 1. It obeys the Law of Conservation of Mass, ensuring that the mass of reactants equals the mass of products. 2. It provides stoichiometric coefficients, which are crucial for carrying out stoichiometry calculations. 3. It helps predict the reactant amounts required to produce a desired quantity of products, as well as the product amounts formed from given quantities of reactant. 4. It allows for efficient planning and optimization of chemical reactions, as well as reducing waste by accurately calculating the required amounts of reactants. By using balanced chemical equations, we ensure the accuracy of our calculations related to chemical reactions and their material quantifications, making it essential for determining the quantity of a product formed from a given quantity of a reactant.

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Most popular questions from this chapter

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) together with other substances. Reaction of the ore with CO produces iron metal: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{CO}_{2}(g) $$ (a) Balance this equation. (b) Calculate the number of grams of CO that can react with \(0.150 \mathrm{~kg}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) (c) Calculate the number of grams of Fe and the number of grams of \(\mathrm{CO}_{2}\) formed when \(0.150 \mathrm{~kg}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of \(\mathrm{CO}, 0.733 \mathrm{~g}\) of \(\mathrm{CO}_{2}\), and \(0.450 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

Determine the empirical and molecular formulas of each of the following substances: (a) Styrene, a compound substance used to make Styrofoam \(^{8}\) cups and insulation, contains \(92.3 \% \mathrm{C}\) and \(7.7 \%\) H by mass and has a molar mass of \(104 \mathrm{~g} / \mathrm{mol}\). (b) Caffeine, a stimulant found in coffee, contains \(49.5 \%\) \(\mathrm{C}, 5.15 \% \mathrm{H}, 28.9 \% \mathrm{~N}\), and \(16.5 \% \mathrm{O}\) by mass and has a molar mass of \(195 \mathrm{~g} / \mathrm{mol}\). (c) Monosodium glutamate (MSG), a flavor enhancer in certain foods, contains \(35.51 \%\) C, \(4.77 \%\) H, \(37.85 \%\) O, \(8.29 \% \mathrm{~N}\), and \(13.60 \% \mathrm{Na}\), and has a molar mass of \(169 \mathrm{~g} / \mathrm{mol}\).

A piece of aluminum foil \(1.00 \mathrm{~cm}\) square and \(0.550 \mathrm{~mm}\) thick is allowed to react with bromine to form aluminum bromide as shown in the accompanying photo. (a) How many moles of aluminum were used? (The density of aluminum is \(2699 \mathrm{~g} / \mathrm{cm}^{3}\) ) (b) How many grams of aluminum bromide form, assuming the aluminum reacts completely?

Balance the following equations, and indicate whether they are combination, decomposition, or combustion reactions: (a) \(\mathrm{Al}(\mathrm{s})+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{AlCl}_{3}(\mathrm{~s})\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{Li}(s)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Li}_{3} \mathrm{~N}(s)\) (d) \(\mathrm{PbCO}_{3}(s) \longrightarrow \mathrm{PbO}(s)+\mathrm{CO}_{2}(g)\) (e) \(\mathrm{C}_{7} \mathrm{H}_{8} \mathrm{O}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

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