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What do the coefficients of a balanced chemical equation tell us about the proportions in which atoms and molecules react on an individual (microscopic) basis?

Short Answer

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The coefficients in a balanced chemical equation indicate the proportions in which atoms and molecules react on a microscopic basis. They show how the reactants combine in specific ratios to form the products, and these ratios remain constant whether we consider individual molecules or moles of the substances involved in the reaction.

Step by step solution

01

Understanding a balanced chemical equation

A balanced chemical equation is a representation of a chemical reaction where the number of atoms of each element on the reactant side is equal to the number of atoms of the same element on the product side. This follows the Law of Conservation of Matter, which states that matter cannot be created or destroyed in a chemical reaction.
02

Interpreting coefficients

In a balanced chemical equation, the numbers in front of the chemical formulas (called coefficients) represent the number of molecules (or moles) that participate in the reaction. For example, in the equation: \(2H_2 + O_2 \rightarrow 2H_2O\) The coefficients are 2 for \(H_2\), 1 for \(O_2\), and 2 for \(H_2O\). These coefficients show that 2 molecules of hydrogen gas react with 1 molecule of oxygen gas to produce 2 molecules of water.
03

Proportions of atoms and molecules at a microscopic level

The coefficients in a balanced chemical equation give us the proportions of atoms and molecules involved in the reaction at a microscopic level. In the example above, we see that 2 hydrogen molecules react with 1 oxygen molecule to form 2 water molecules. This means that the ratio of hydrogen to oxygen molecules is 2:1, and the ratio of hydrogen to water molecules is also 2:1. In other words, the coefficients tell us the ratio in which the reactants combine to form the products.
04

Coefficients can also apply to moles

The coefficients in a balanced chemical equation can also be interpreted as the number of moles of each substance. One mole of a substance contains Avogadro's number (\(6.022 \times 10^{23}\)) of particles (atoms, ions, molecules, etc.). In this context, instead of counting individual molecules, we count moles. For example: \(2 \, moles \, H_2 + 1 \, mole \, O_2 \rightarrow 2 \, moles \, H_2O\) This means that the ratio of moles of hydrogen to moles of oxygen is still 2:1, and the ratio of moles of hydrogen to moles of water is also 2:1. In conclusion, the coefficients in a balanced chemical equation indicate the proportions in which atoms and molecules react on a microscopic basis. They show how the reactants combine in specific ratios to form the products, and these ratios remain constant whether we consider individual molecules or moles of the substances involved in the reaction.

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Most popular questions from this chapter

For each of the following unbalanced equations, suppose that exactly \(1.00 \mathrm{g}\) of each reactant is taken. Determine which reactant is limiting, and calculate what mass of the product in boldface is expected (assuming that the limiting reactant is completely consumed). a. \(\mathrm{CS}_{2}(l)+\mathrm{O}_{2}(g) \rightarrow \mathbf{C} \mathbf{O}_{2}(g)+\mathrm{SO}_{2}(g)\) b. \(\mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CN}_{2} \mathrm{H}_{4} \mathrm{O}(s)+\mathbf{H}_{2} \mathbf{O}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{MnO}_{2}(s) \rightarrow \mathrm{MnO}(s)+\mathbf{H}_{2} \mathbf{O}(g)\) d. \(\mathrm{I}_{2}(l)+\mathrm{Cl}_{2}(g) \rightarrow \mathbf{I C l}(g)\)

One method for chemical analysis involves finding some reagent that will precipitate the species of interest. The mass of the precipitate is then used to determine what mass of the species of interest was present in the original sample. For example, calcium ion can be precipitated from solution by addition of sodium oxalate. The balanced equation is $$\mathrm{Ca}^{2+}(a q)+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \rightarrow \mathrm{CaC}_{2} \mathrm{O}_{4}(s)+2 \mathrm{Na}^{+}(a q)$$ Suppose a solution is known to contain approximately 15 g of calcium ion. Show by calculation whether the addition of a solution containing \(15 \mathrm{g}\) of sodium oxalate will precipitate all of the calcium from the sample.

Using the average atomic masses given inside the front cover of the text, calculate the mass in grams of each of the following samples. a. 5.0 mol of nitric acid b. 0.000305 mol of mercury c. \(2.31 \times 10^{-5}\) mol of potassium chromate d. 10.5 mol of aluminum chloride e. \(4.9 \times 10^{4}\) mol of sulfur hexafluoride f. 125 mol of ammonia g. 0.01205 mol of sodium peroxide

The more reactive halogen elements are able to replace the less reactive halogens from their compounds. $$\begin{array}{l} \mathrm{Cl}_{2}(g)+\mathrm{NaI}(a q) \rightarrow \mathrm{NaCl}(a q)+\mathrm{I}_{2}(s) \\ \mathrm{Br}_{2}(l)+\mathrm{NaI}(a q) \rightarrow \mathrm{NaBr}(a q)+\mathrm{I}_{2}(s) \end{array}$$ Suppose separate solutions each containing \(25.0 \mathrm{g}\) of NaI are available. If \(5.00 \mathrm{g}\) of \(\mathrm{Cl}_{2}\) gas is bubbled into one NaI solution, and \(5.00 \mathrm{g}\) of liquid bromine is added to the other, calculate the number of grams of elemental iodine produced in each case.

For each of the following unbalanced chemical equations, suppose that exactly \(5.00 \mathrm{g}\) of each reactant is taken. Determine which reactant is limiting, and calculate what mass of each product is expected (assuming that the limiting reactant is completely consumed). a. \(\mathrm{S}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) b. \(\operatorname{MnO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \rightarrow \mathrm{Mn}\left(\mathrm{SO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) c. \(\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) d. \(\mathrm{AgNO}_{3}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{Ag}(s)+\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)\)

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