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(a) Define the terms limiting reactant and excess reactant. (b) Why are the amounts of products formed in a reaction determined only by the amount of the limiting reactant? (c) Why should you base your choice of what compound is the limiting reactant on its number of initial moles, not on its initial mass in grams?

Short Answer

Expert verified
(a) The limiting reactant is the reactant in a chemical reaction that is completely consumed when the reaction reaches completion, determining the maximum amount of product that can be formed. The excess reactant is the reactant that is not completely used up in the reaction and is present in a higher amount compared to the stoichiometric ratio needed. (b) The amounts of products formed in a reaction are determined only by the amount of the limiting reactant because once the limiting reactant is depleted, the reaction stops, even if there are excess reactants left, as there is nothing left for the excess reactants to react with. (c) The choice of limiting reactant should be based on its number of initial moles, not on its initial mass in grams because the stoichiometry of a reaction is based on a mole-to-mole ratio of reactants and products, not on mass ratios. Comparing the number of moles of each reactant present with their stoichiometric ratio helps determine which reactant will be used up first. Using mass instead of moles could lead to incorrect conclusions about the limiting reactant, as it doesn't take into account the different molar masses of the compounds involved in the reaction.

Step by step solution

01

(a) Define the terms limiting reactant and excess reactant.

The limiting reactant is the reactant in a reaction that is completely consumed and determines the maximum amount of product that can be formed. The excess reactant is the reactant that isn't completely used up in the reaction and is present in excess compared to the stoichiometric ratio needed. In other words, there is more of the excess reactant available than is needed for the reaction to complete.
02

(b) Why are the amounts of products formed in a reaction determined only by the amount of the limiting reactant?

The amounts of products formed in a reaction are determined only by the amount of the limiting reactant because the limiting reactant is the one that will be completely consumed to produce the maximum possible amount of the product. Once the limiting reactant is depleted, the reaction stops even if there is an excess of other reactants, as there will be nothing left for the excess reactants to react with.
03

(c) Why should you base your choice of what compound is the limiting reactant on its number of initial moles, not on its initial mass in grams?

You should base your choice of what compound is the limiting reactant on its number of initial moles and not on its initial mass in grams because the stoichiometry of a reaction is based on a mole-to-mole ratio of reactants and products, not on mass ratios. By comparing the number of moles of each reactant present with their stoichiometric ratio, you can determine which reactant will be used up first and, therefore, which one is the limiting reactant. Using mass instead of moles could lead to incorrect conclusions about the limiting reactant, as it doesn't take into account the different molar masses of the compounds involved in the reaction.

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Most popular questions from this chapter

Hydrogen sulfide is an impurity in natural gas that must be removed. One common removal method is called the Claus process, which relies on the reaction: $$ 8 \mathrm{H}_{2} \mathrm{~S}(g)+4 \mathrm{O}_{2}(g) \longrightarrow \mathrm{S}_{8}(l)+8 \mathrm{H}_{2} \mathrm{O}(g) $$ Under optimal conditions the Claus process gives \(98 \%\) yield of \(\mathrm{S}_{8}\) from \(\mathrm{H}_{2} \mathrm{~S}\). If you started with 300 grams of \(\mathrm{H}_{2} \mathrm{~S}\) and \(50.0\) grams of \(\mathrm{O}_{2}\), how many grams of \(\mathrm{S}_{\mathrm{g}}\) would be produced, assuming \(98 \%\) yield?

Balance the following equations: (a) \(\mathrm{L}_{\mathrm{L}(s)}+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Li}_{3} \mathrm{~N}(s)\) (b) \(\mathrm{La}_{2} \mathrm{O}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{La}(\mathrm{OH})_{3}(a q)\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{PH}_{3}(g)\) (e) \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q)\) \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (f) \(\mathrm{AgNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \overrightarrow{\mathrm{Ag}_{2} \mathrm{SO}_{4}(s)+\mathrm{NaNO}_{3}(a q)}\) (g) \(\mathrm{CH}_{3} \mathrm{NH}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow\) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)\)

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(3.0 \times 10^{23}\) molecules of \(\mathrm{H}_{2} \mathrm{O}_{2}, 2.0 \mathrm{~mol} \mathrm{CH}_{4}, 32 \mathrm{~g} \mathrm{O}_{2}\).

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{CH}_{2}\), molar mass \(=84 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{NH}_{2} \mathrm{Cl}\), molar mass \(=51.5 \mathrm{~g} / \mathrm{mol}\)

An oxybromate compound, \(\mathrm{KBrO}_{x}\), where \(x\) is unknown, is analyzed and found to contain \(52.92 \%\). Br. What is the value of \(x ?\)

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