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Chapter 9: Problem 12

Describe a general plan for solving all stoichiometry problems in three steps.

Short Answer

Expert verified
To solve stoichiometry problems generally involves: 1) Converting given quantities to moles; 2) Using the balanced chemical equation to relate quantities of one substance to another using mole ratios; 3) Converting the obtained quantity in moles back into required units.

Step by step solution

01

Convert all given quantities to moles

Firstly, convert all the given quantities of substances into moles. This is because stoichiometry problems are best solved in the realm of moles. Use the molar mass of each substance to convert from grams to moles. The molar mass of a substance is the mass in grams of one mole of that substance. It can be calculated using the atomic masses of the elements that comprise the substance. If the quantity is given in a volume (in liters), then use the ideal gas law equation PV=nRT to find the number of moles n. In this equation, P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.
02

Use the balanced chemical equation

The balanced chemical equation gives the relationship between the reactants and the products and can provide the moles of a given substance involved in the reaction. This is represented in the form of a mole ratio and provides a bridge from one substance to another. In other words, it tells you how many moles of one substance react or form in relation to the other substance. Use the coefficients of the reactants and products in the balanced chemical equation to set up a conversion factor.
03

Convert moles to desired units

The third step involves converting the moles of the substance of interest to its required units, as per the demands of the problem. This could be back into grams or volumes. In order to convert moles to grams, use the molar mass of the substance. Multiply the number of moles by the molar mass. For gases, use the ideal gas law equation \(PV=nRT\) to convert from moles to a required volume.

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Most popular questions from this chapter

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

Why is a balanced chemical equation needed to solve stoichiometry problems?

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 \(\mathrm{g} / \mathrm{mL} ?\)

Use the equation provided to answer the questions that follow. The density of oxygen gas is 1.428 \(\mathrm{g} / \mathrm{L}\) . $$2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)$$ \begin{equation}\begin{array}{l}{\text { a. What volume of oxygen can be made }} \\ {\text { from } 5.00 \times 10^{-2} \text { mol of } \mathrm{KClO}_{3} ?} \\ {\text { b. How many grams } \mathrm{KClO}_{3} \text { must react to }} \\\ {\text { form } 42.0 \mathrm{mL} \mathrm{O}_{2} \text { ? }} \\ {\text { c. How many milliliters of } \mathrm{O}_{2} \text { will form at }} \\ {\text { STP from } 55.2 \mathrm{g} \mathrm{KClO}_{3} ?}\end{array}\end{equation}

Define stoichiometry.
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