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Chapter 9: Problem 56

Use the following terms to create a concept map: stoichiometry, excess reactant, theoretical yield, and mole ratio.

Short Answer

Expert verified
A concept map starts with stoichiometry at the centre, which is directly connected with the excess reactant, theoretical yield, and mole ratio. Theoretical yield is also connected to the mole ratio, which shows its dependency on it. Similarly, excess reactant and theoretical yield too are interconnected.

Step by step solution

01

Understanding Key Terms

It's important to first understand each term. Stoichiometry is the relationship between the reactants and products in a chemical reaction. Excess reactant is the reactant in a chemical reaction that remains after the reaction stops. Theoretical yield is the maximum amount of product that could be created from a given amount of reactant. Mole ratio is the ratio of moles of one substance to the moles of another substance in a balanced chemical equation.
02

Creating the Concept Map

Put 'Stoichiometry' in the center of the map. This is because all the other concepts are part of stoichiometry. Draw lines from stoichiometry to the other terms. For instance, 'excess reactant' and 'theoretical yield' are directly dependent on stoichiometry in chemical reactions, so you can connect these terms directly to stoichiometry. Similarly, 'mole ratio' can be connected directly to 'stoichiometry'. Now, connect 'theoretical yield' with 'mole ratio'. This indicates that theoretical yield is calculated using the mole ratio. The connection between 'excess reactant' and 'theoretical yield' denotes that the excess reactant determines the theoretical yield.
03

Review and Verify

After creating the concept map, cross-check to ensure all concepts are accurately represented and connected. If a term is not properly placed or connected, rectify it to show the correct relationship.

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Most popular questions from this chapter

Write the conversion factor needed to convert from g \(\mathrm{O}_{2}\) to \(\mathrm{L} \mathrm{O}_{2}\) if the density of \(\mathrm{O}_{2}\) is 1.429 \(\mathrm{g} / \mathrm{L}\) .

Calcium carbide, \(\mathrm{CaC}_{2},\) reacts with water to form acetylene. $$\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s)$$ \begin{equation}\begin{array}{l}{\text { a. How many grams of water are needed to }} \\ {\text { react with } 485 \text { g of calcium carbide? }} \\\ {\text { b. How many grams of } \mathrm{CaC}_{2} \text { could make }} \\\ {\text { 23.6 g } \mathrm{C}_{2} \mathrm{H}_{2} \text { ? }} \\ {\text { c. If } 55.3 \mathrm{g} \mathrm{Ca}(\mathrm{OH})_{2} \text { are formed, how many }} \\ {\text { grams of water reacted? }}\end{array}\end{equation}

If 156 \(\mathrm{g}\) of sodium nitrate react, and 112 \(\mathrm{g}\) of sodium nitrite are recovered, what is the percentage yield? $$2 \mathrm{NaNO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{2}(s)+\mathrm{O}_{2}(g)$$

How many liters of \(\mathrm{O}_{2},\) density \(1.43 \mathrm{g} / \mathrm{L},\) are needed for the complete combustion of \(1.00 \mathrm{L} \mathrm{C}_{8} \mathrm{H}_{18},\) density 0.700 \(\mathrm{g} / \mathrm{mL} ?\)

How many grams of Al form from 9.73 g of aluminum oxide if the yield is 91\(\% ?\) $$\mathrm{Al}_{2} \mathrm{O}_{3}+3 \mathrm{C} \rightarrow 2 \mathrm{Al}+3 \mathrm{CO}$$
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