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Chapter 9: Problem 6

Why is it necessary to use mole ratios in solving stoichiometry problems?

Short Answer

Expert verified
Mole ratios are necessary in stoichiometry because they allow us to calculate the relationship between the reactants and products in a chemical reaction. This is crucial when predicting the amounts of reactants necessary or product produced in a reaction.

Step by step solution

01

Define Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships of the reactants and products in a chemical reaction.
02

Explain Mole Ratio

Mole ratio refers to the ratio of moles of one substance to the moles of another substance in a balanced chemical equation.
03

Relate Mole Ratio and Stoichiometry

Mole ratios are used in stoichiometry calculations as a conversion factor. In a balanced chemical equation, the coefficients represent the ratio of the moles involved in the reaction. Hence, it allows us to calculate how much product will be formed from a certain amount of reactant or how much reactant is needed to produce a certain amount of product.
04

Conclude on the Necessity of Mole Ratios in Stoichiometry

Therefore, without knowing the mole to mole ratio, we can't accurately determine amounts of reactants or products involved, making it impossible to solve stoichiometric problems.

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Most popular questions from this chapter

Give two reasons why the actual yield from chemical reactions is less than 100\(\% .\)

How many grams of \(\mathrm{NaNO}_{2}\) form when 256 \(\mathrm{g} \mathrm{NaNO}_{3}\) react? The yield is 91\(\% .\) $$2 \mathrm{NaNO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{2}(s)+\mathrm{O}_{2}(g)$$

What is the function of the catalytic converter in the exhaust system?

The percentage yield of nitric acid is 95\(\% .\) If 9.88 \(\mathrm{kg}\) of nitrogen dioxide react, what mass of nitric acid is isolated? $$3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)$$

Graphing Calculator Calculating Percentage Yield of a Chemical Reaction The graphing calculator can run a program that calculates the percentage yield of a chemical reaction when you enter the actual yield and the theoretical yield. Using an example in which the actual yield is 38.8 g and the theoretical yield is 53.2 g, you will calculate the percentage yield. First, the pro- gram will carry out the calculation. Then you can use it to make other calculations. Go to Appendix C. If you are using a TI-83 Plus, you can download the program YIELD and data and run the application as directed. If you are using another calculator, your teacher will provide you with keystrokes and data sets to use. After you have run the program, answer the questions. Note: all answers are written with three significant figures. \begin{equation}\begin{array}{l}{\text { a. What is the percentage yield when the }} \\ {\text { actual yield is } 27.3 \mathrm{g} \text { and the theoretical }} \\ {\text { yield is } 44.6 \mathrm{g} \text { ? }} \\ {\text { b. What is the percentage yield when the }} \\ {\text { actual yield is } 5.40 \mathrm{g} \text { and the theoretical }} \\ {\text { c. What actual yield theoretical yield pair }} \\ {\text { produced the largest percentage yield? }}\end{array}\end{equation}
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