Chemical equations mole ratios from

To solve the problem, you need to know how the unknown moles of hydrogen are related to the known moles of potassium. In Section 11.1, you learned to derive mole ratios from the balanced chemical equation. Mole ratios are used as conversion factors to convert the known number of moles of one substance to the unknown number of moles of another substance in the same reaction. Several mole ratios can be written from the equation, but how do you choose the correct one ... 

Why learn to write mole ratios They are the key to calculations that are based on chemical equations. Using a balanced chemical equation, mole ratios derived from the equation, and a given amount of one of the reactants or products, you can calculate the amount of any other participant in the reaction. 

Step 2 Use the mole ratio derived from the stoichiometric coefficients in the balanced chemical equation to convert from the amount of one substance (A) into the amount in moles of the other substance (B). For aA - / B or aA + hY> — cC, use... 

In this approach, we use the mole ratio from the chemical equation to determine whether there is enough of one reactant to react with another. 

To summarize, the amounts of different reagents that participate in a chemical reaction are related through the stoichiometric coefficients in the balanced chemical equation. To convert from moles of one reagent to moles of any other reagent, multiply by the stoichiometric ratio that leads to proper cancellation of units ... 

Convert the masses of the reactants and products to moles using their molar masses. Using the mole ratios from the balanced chemical equation, it is possible to determine how much material should react or be produced. These calculated values can be compared to the observed values. 

The balanced chemical equation for this reaction shows that 1 mol of nitric acid reacts with 1 mol of sodium hydroxide. If equal molar quantities of nitric acid and sodium hydroxide are used, the result is a neutral (pH 7) aqueous solution of sodium nitrate. In fact, when any strong acid reacts with any strong base in the mole ratio from the balanced chemical equation, a neutral aqueous solution of a salt is formed. Reactions between acids and bases of different strengths usually do not result in neutral solutions. 

At the beginning of this chapter, you were introduced to Gay-Lussac s law of combining volumes When gases react, the volumes of the reactants and the products, measured at equal temperatures and pressures, are always in whole number ratios. As well, you learned that the mole ratios from a chemical equation are the same as the ratios of the volumes of the gases. 

Use proportional reasoning to determine mole ratios from a balanced chemical equation. 

Substances are usually measured by mass or volume. As a result, before using the mole ratio you will often need to convert between the units for mass and volume and the unit mol Yet each stoichiometry problem has the step in which moles of one substance are converted into moles of a second substance using the mole ratio from the balanced chemical equation. Follow the steps in Skills Toolkit 2 to understand the process of solving stoichiometry problems. 

You will write mole ratios from balanced chemical equations. 

Determine the moles of the unknown substance from the moles of the given substance. Use the appropriate mole ratio from the balanced chemical equation as the conversion factor. 

We express all concentrations in moles per liter. The mole ratio from the balanced chemical equation allows us to find the changes in concentrations of the other substances in the reaction. We use the reaction summary to find the equihbrium concentrations to use in the... 

A titration problem is an applied stoichiometry problem, so we will need a balanced chemical equation. We know the molarity and volume for the NaOH solution, so we can find the number of moles reacting. The mole ratio from the balanced equation lets us calculate moles of H2SO4 firom moles of NaOH. Because we know the volume of the original H2SO4 solution, we can find its molarity. 

This is the type of problem that smdents get wrong if they use the dilution formula (Eq. 3.3) by mistake. The dilution formula does not include the mole ratio from a balanced chemical equation, and in this case, the 2 1 ratio would result in an error of a factor of 2 if you used Eq. 3.3. 

You are starting with moles of iron and want grams of Fe203 so we ll first convert from moles of iron to moles of Fe203 using the ratio of moles of Fe203 to moles of iron as defined by the balanced chemical equation ... 

The mole is the most important concept in this chapter. Nearly every problem associated with this material requires moles in at least one of the steps. You should get into the habit of automatically looking for moles. There are several ways of finding the moles of a substance. You may determine the moles of a substance from a balanced chemical equation. You may determine moles from the mass and molecular weight of a substance. You may determine moles from the number of particles and Avogadro s number. You may find moles from the moles of another substance and a mole ratio. Later in this book, you will find even more ways to determine moles. In some cases, you will be finished when you find moles, in other cases, finding moles is only one of the steps in a longer problem. 

The term (2 mol HC1/1 mol H2) is a mole ratio. We got this mole ratio directly from the balanced chemical equation. The balanced chemical equation has a 2 in front of the HC1, thus we have the same number in front of the mol HC1. The balanced chemical equation has an understood 1 in front of the H2, for this reason the same value belongs in front of the mol H2. The values in the mole ratio are exact numbers, and, as such, do not affect the significant figures. 

To find the moles of IF5 from the limiting reagent, we need to use a mole ratio derived from information in the balanced chemical equation. (This is another place where, if we had not balanced the equation, we would be in trouble.)... 

You should be very careful when working problems involving gases and one or more other phases. The gas laws can only give direct information about gases. This is why there is a mole ratio conversion (from the balanced chemical equation) in this example to convert from the solid (KCI03) to the gas (02). 

Be able to convert from moles of one substance to moles of another, using the stoichiometric ratio derived from the balanced chemical equation. 

The pipeted volume is converted to moles by multiplying the liters of solution by its molarity. The moles of titrant are determined from the mole ratio in the balanced chemical equation for the reaction. The molarity of the solution is calculated by dividing the moles of titrant by the liters of titrant used. 

The balanced chemical equation for a reaction is used to set up the conversion factor from one substance to another and that conversion factor, the mole ratio for the reaction, is applied to the moles given to calculate the moles required. 

To interpret a titration, we need the stoichiometric relation from the chemical equation for the reaction. This relation is used to write the mole ratio in the usual way. The only new step is to use the molarities of the solutions to convert between the moles of reactants and the volumes of... 

The most important step to all of these calculations is the use of a value known as the mole ratio. The mole ratio is the ratio of moles of one substance to moles of second substance. It is determined by the ratios of the coefficients from the balanced chemical equation. The mole ratio is used in all conversions since it allows you to switch from values that describe the given substances to values that describe the unknown substance. To facilitate this process, there is another chart, Figure 12.2, that provides guidelines for solving most problems. In this first type of calculation, we will use the mole ratio to convert from units of moles of the given substance to moles of the unknown substance. We re going to omit the states of the reactants and products so that you can focus your attention on the coefficients. 

It s fairly easy to conceptualize the idea of limiting reactants when you are given moles of the reactants. When you are given grams, it is not always so easy to see. When you have to solve limiting reactant problems, it is always necessary to determine the number of moles of each substance and compare that to the required ratios from the balanced chemical equation. Let s use the same reaction, but use masses instead of moles. 

A mole ratio is the ratio of the number of moles of one substance to the number of moles of another substance. Because coefficients can represent moles, molecules, or atoms, you can think of a mole ratio as a coefficient ratio. For example, look at the equation for aluminum oxide, 4A1 + 302 —> 2Al203. You can pick any two substances from the equation and determine their mole ratio. The mole ratio of Al to Al203 is 4 2 or 2 1, while the mole ratio of 02 to Al203 is 3 2. This leads to another type of problem that you might encounter. Suppose you were asked to produce 1,000 moles of Al203 for a big chemical company. How much aluminum and oxygen would you need to purchase Start with the balanced equation 4A1 + 302 — 2Al203. 

The coefficients in a balanced equation give the ratio of moles of each substance in the reaction to moles of any other substance. They also give the ratio of formula units of each substance to formula units of any other substance. The balanced chemical equation is the cornerstone from which we can calculate how much of one substance reacts with or is produced by a certain quantity of another substance (Chapter 10). 

Inst as compounds have definite ratios of elements, chemical reactions have definite ratios of reactants and products. Those ratios are used in Section 10.1 to calcnlate the number of moles of other substances in a reaction from the nnm-ber of moles of any one of the snbstances. Section 10.2 combines information from Section 10.1, Chapter 7, and elsewhere to explain how to calcnlate the mass of any substance involved in a reaction from the mass of another. Section 10.3 demonstrates how to work with qnantities in nnits other than moles or masses when finding quantities of reactants or prodncts. Section 10.4 shows how to calcnlate the quantities of snbstances involved in a reaction even if the quantities of reactants present are not in the mole ratio of the balanced equation. Section 10.5 covers the calculation of the percentage yield of a product from the actual yield and the theoretical yield, based on the amonnt(s) of reactant(s). Section 10.6 explains which of these types of calcnlations can and cannot be done with net ionic equations. 

It must be noted that the magnimdes of the quantities in the change due to reaction line are always in the ratio of the coefficients in the balanced chemical equation. It will also become apparent that the numbers of moles of reactants in the change line are subtracted from the initial quantities present and the numbers of moles of products are added to any initial quantities present. 

Step 4 Complete the change row by writing the number of moles of each substance that would react with or be produced from the quantity in step 3. Use a minus sign with each quantity of reactant. The magnitudes in the change row are in the same ratio as the coefficients in the balanced chemical equation. 

Convert the moles of H2SO4 to moles of (NH4)2S04, using the ratio from the balanced chemical equation. Convert the moles of (NH4)2S04 to mass of (NH4)2S04, using its molar mass. 

To determine the concentration of a solute in a solution (e.g., HCl) we treat the unknown solution with a solution of known concentration and volume (e.g., NaOH) until the mole ratio is exactly what is required by the balanced chemical equation. Then from the known volumes of both solutions, the unknown concentration of the... 

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