### SAT Chemistry Stoichiometry (Chemical Calculations) and the Mole Concept - Stoichiometry: Mole-Mole Problems

##
**SAT Chemistry Stoichiometry (Chemical Calculations) and the Mole Concept - Stoichiometry: Mole-Mole Problems**

The types of mole problems investigated so far have been ones involving only one substance. Chemical calculations often involve more than one substance and take into consideration information found in balanced reaction equations discussed in Chapter 4. Recall that coef¬ficients from a balanced equation can be used to describe the numbers of atoms, ions, or molecules involved in the chemical process. Also recall that the quantities of each must balance on both sides of the equation to satisfy the Law of Conservation of Matter. Those coefficients can also represent larger numbers of particles, namely moles of those substances, that are reacting or being produced. Therefore, the balanced chemical reaction2NaClO(s) → 2NaCl(s) + O

_{2}(g)

can be interpreted in two ways.

**1.**Decomposing 2 formula units of sodium hypochlorite produces 2 formula units of sodium chloride and 1 molecule of oxygen.

**2.**Decomposing 2 moles of sodium hypochlorite produces 2 moles of sodium chloride and 1 mole of oxygen molecules.

Coefficients in balanced chemical reaction equations therefore provide mole ratios for react¬ing substances and substances produced.

**Example 1**__________________________________________

How many moles of sodium chloride can be produced from 0.0253 moles of sodium hypochlorite?

Use dimensional analysis:

**Example 2**__________________________________________

How many moles of sodium hypochlorite are needed to produce 0.750 moles of oxygen?

Use dimensional analysis:

**STOICHIOMETRY: MASS-MASS PROBLEMS**

In order to work with substances in the laboratory, chemists must work with quantities they can easily measure. A mole is an amount impractical to actually count out in the lab. That’s why the concept of molar mass, which relates moles to mass, was discussed in a previous section of this chapter. Using molar mass allows problems to be based on mass, which is a measurable quantity. Mass-mass problems often involve determining the masses of other substances needed to react with a given mass of a substance or the mass of other substances that can be produced from that given mass.

**Example**__________________________________________

Consider the following balanced chemical reaction equation:

CaCO

_{3}(s) + 2HCl(aq) —> CaCl

_{2}(aq) + H

_{2}O(1) + CO

_{2}(g)

If 2.59 g of CaCO

_{3}was reacted with enough HC1 to use up all of the CaCO

_{3}, what mass of HCl would be needed and what mass of CO

_{2}would be produced?

Use dimensional analysis:

The first factor in the dimensional analysis equation converts the mass of the CaCO

_{3}to moles by using information concerning molar mass as found from the Periodic Table. The second factor converts moles of CaCO

_{3}to moles of HCl needed to react with the CaCO

_{3}. The mole ratio was found from the coefficients in front of each substance in the balanced reaction. The third factor converts moles of HCl to grams of HCl, once again using molar mass as found from the Periodic Table.

To find the mass of CO

_{2}produced, set up a similar dimensional analysis equation. First convert the mass of CaCO

_{3}to moles. This time, however, use the mole ratio for CO

_{2}and CaCO

_{3 }for the second factor. Finally, use the molar mass of CO

_{2}to convert moles of CO

_{2}to grams

**STOICHIOMETRY: VOLUME-VOLUME PROBLEMS**

In the lab, gases are often used and their volumes are easily measured. Recall that the molar volume of any gas at STP is 22.4 L. So the volume of a gas at standard temperature and pres¬sure can be converted to moles of that gas via dimensional analysis. Combining those conver¬sions with the mole ratios found in balanced equations, for reactions involving gases, allows for volume-volume stoichiometry problems to be calculated. In volume-volume problems, you are given the volume of one gas at STP and asked to determine the volume (s) of other gases involved in the reaction.

**Example**__________________________________________

Consider the following balanced chemical reaction equation:

N

_{2}(g) + 3H

_{2}(g) -> 2NH

_{3}(g)

To produce 0.400 L of NH

_{3}at STP, what volumes of nitrogen and hydrogen, also at STP, would be required?

Use dimensional analysis:

The first factor in the dimensional analysis equation converts the volume of NH

_{3}to moles. The second factor uses the mole ratio from the balanced equation to convert moles of NH

_{3 }to moles of N

_{2}. Finally, the third factor converts moles of N

_{2}to the volume of N

_{2}. Mathematically, factors #1 and #3 simply undo each other. They display a relationship that was suggested in Chapter 5 by the Ideal Gas Law as well as by Avogadro’s Hypothesis (discussed in this chapter). Namely, there is a direct relationship between the volumes of gases, at the same temperature and pressure, and their numbers of particles (measured in moles). In other words, mole ratios can be construed as volume ratios between gases existing at the same temperature and pressure. The only factor needed to solve the previous problem mathematically was factor #2. To express the dimensional analysis equation properly, though, requires using the mole ratios as volume ratios as is done here:

To find the amount of H

_{2}required to produce the 0.400 L of NH

_{3}requires a similar equation but with a different ratio between the gases:

Using mole ratios, i.e., the coefficients from the balanced equation, as volume ratios saves time. It also makes volume-volume problems less cumbersome to solve. The relationship between the volumes of reacting gases was first noted by the French scientist Joseph Louis Gay-Lussac and is sometimes called Gay-Lussac’s Law of Combining Gases. This law states

that when only gases are involved in a chemical reaction, the volumes of the reacting gases and the volumes of the gaseous products are in small whole-number ratios with each other. Those small whole numbers are the coefficients in the balanced reaction equation.

Often reactions between gases do not occur at STP. However, Gay-Lussac’s Law still applies. The reason is fundamentally due to Avogadro’s Law. That law shows that the only requirement for the volumes of gases to be related to the number of particles of those gases is that the temperature and pressure of the gases be the same. Whether or not the temperature and pressure of the gases are at STP is inconsequential. Using the coefficients from the bal¬anced reaction equation as volume ratios between reacting gases in dimensional analysis/stoichiometry problems is encouraged.

**STOICHIOMETRY: MASS-VOLUME OR VOLUME-MASS PROBLEMS**

Reactions often involve gases and other phases of matter. In those reactions, it is common to know the mass of one substance involved in the chemical process and the need to determine the volume of a different substance, such as a gas. Likewise, it is not unusual to know the volume of a gas taking part in a reaction and the need to determine the mass of another substance, often a solid or liquid. Even if the reaction is taking place at STP, Gay-Lussac’s Law cannot be taken advantage of here since both of the substances are not gases and the information desired is not restricted to just volumes. In other words, Gay-Lussac’s Law applies only when all the substances being considered are gases.

**Example 1**__________________________________________

In the reaction below, what mass of magnesium is required to produce 0.250 L of H

_{2}at STP?

Mg(s) + 2HCl(aq) -> MgCl

_{2}(aq) + H

_{2}(g)

The solution to this problem uses both molar mass and molar volume. Use dimensional analysis:

Reactions involving gases and other phases of matter NOT at STP are very common. Obvi-ously, Gay-Lussac’s Law cannot be used to solve such problems since both of the substances in question are not gases. Additionally, the relationship 22.4 L = 1 mole of the gas cannot be used since the reaction is not taking place at STP To solve problems such as these, the Ideal Gas Law, PV = nRT, must be considered. Recall from Chapter 5 that the Ideal Gas Law can be manipulated to solve for the moles of gas at any temperature and pressure as long as the volume is supplied (n = PV/RT). of moles of the gas is known along with its temperature and pressure (V = nRT/P).

**Example 2**__________________________________________

Based on the reaction below, what volume of oxygen is produced from the decomposition of

5.0 g of hydrogen peroxide if the oxygen produced was collected at 70.0°C and 1.25 atm pressure?

2H

_{2}O

_{2}(aq) —> 2H

_{2}O(l) + O

_{2}(g)

To solve this problem, dimensional analysis should be used to determine the number of moles of O

_{2}that would be produced:

The number of moles of O

_{2}produced can now be plugged into the Ideal Gas Law to find the volume of that amount of O

_{2}at the temperature and pressure outlined in the problem:

**PROBLEMS WITH AN EXCESS OF ONE REACTANT OR A LIMITING REACTANT**

It will not always be true that the amounts given in a particular problem are exactly in the proportion required for the reaction to use up all of the reactants. In other words, at times some of one reactant will be left over after the other has been used up. This is similar to the situation in which two eggs are required to mix with one cup of flour in a particular recipe, and you have four eggs and four cups of flour.

Since two eggs require only one cup of flour, four eggs can use only two cups of flour and two cups of flour will be left over.

A chemical equation is very much like a recipe.

**Example**__________________________________________

Consider the following reaction:

CH

_{4}(g) + 2O

_{2}(g) → CO

_{2}(g) + 2H

_{2}O(g)

If you are given 15.0 grams of methane (CH

_{4}) and 15.0 grams of oxygen (O

_{2}), how many grams of carbon dioxide gas can be produced? Which reactant will be left over? How much of this reactant will not be used?

Once again, dimensional analysis will be used to solve this problem. However, two equa¬tions will be required as it will be necessary to determine how much carbon dioxide can be produced from each reactant:

Since the oxygen can produce only 10.3 g CO

_{2}(the lesser of the quantities of CO_{2}shown above), it is referred to as the limiting reactant. There simply is not enough of oxygen available to make what the methane has the potential to produce. That smaller amount of CO_{2}is referred to as the theoretical yield. A portion of the CH_{4}(the reactant in excess) will be used, however, to produce the 10.3 g of CO_{2}. To determine that amount, another dimensional analysis is needed.**PERCENT YIELD OF A PRODUCT**

In most stoichiometric problems, we assume that the results are exactly what we would theoretically expect. In reality, the resulting theoretical yield is rarely the actual yield. Why the actual yield of a reaction may be less than the theoretical yield occurs for many reasons. Some of the product is often lost during the purification or collection process.

Chemists are usually interested in the efficiency of a reaction. The efficiency is expressed by comparing the actual and the theoretical yields.

The percent yield is the ratio of the actual yield to the theoretical yield, multiplied by 100%.

**Example**__________________________________________

Aluminum is commonly produced by the smelting of aluminum oxide into aluminum metal by the reaction below:

2Al

_{2}O

_{3}(dissolved) + 3C(s) -> 4Al( ℓ ) + 3CO

_{2}(g)

If 3.89 kg of aluminum oxide is smelted and the actual yield of A1 is 1.95 kg, what is the percent yield associated with the process?

The theoretical yield of aluminum from that amount of aluminum oxide can be found using dimensional analysis:

I really want to thank the author for such a nice blog that helped me to understand why it is important.

ReplyDeletebuy sodium chlorite uk

I found decent information in your article. I am impressed with how nicely you described this subject, It is a gainful article for us. Thanks for share it. Buy sodium chlorite 22.4%

ReplyDelete