CHEM1060 Unit 4→ Chapter 9 Chemical Quantities PDF

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Unit 4→ Chapter 9: Chemical Quantities 9-2: Mole-Mole Relationships Example 9.2 Question: What number of moles of O2 will be produced by the decomposition of 5.8 moles of water? Solution: First, start by writing the balanced equation for the decomposition of water. 2H2O → 2H2 + O2 From this equation, we can state that 2 moles of H2O yields 1 mole of O2 which can be represented by the following equivalence statement, 2 moles H2O = 1 mol O2 We can use this equivalence statement to determine the mole ratio that we need. = 1 mol O2/ 2 moles O2 ← mole ratio Next, we multiply the mole ratio by the # of moles of H2O, 5.8 moles H2O x (1 mol O2/ 2 moles H2O) = 2.9 moles of O2 So, if we decompose 5.8 moles of H2O, we would produce 2.9 moles of O2. Example 9.3 Question: Calculate the number of moles of oxygen required to react exactly with 4.30 moles of propane, C3H8, in the reaction described by the following balanced equation, C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g) Solution: According to the balanced equation, we can state that 1 mol C3H8 requires 5 moles of O2. This can be represented by the following equivalence statement, 1 mol C3H8= 5 moles O2 This leads us to the required mole ratio, 5 moles O2 / 1 mol C3H8 Next, we multiply the mole ratio by the number of moles of propane, C3H8, 4.30 moles C3H8 x (5 moles O2/ 1 mol C3H8) = 21.5 moles O2 So, 4.30 moles of C3H8 requires 21.5 moles of O2 to react completely.

9-3: Mass Calculations In this section, we will review the procedures for converting between moles and masses, and we will see how these procedures are applied to chemical calculations. Example Question

Question: Consider the reaction between aluminum and iodine to produce aluminum iodide. Suppose we have 35.0g of aluminum. What mass of I2 would be required to react with this amount of aluminum? Solution: The balanced equation is provided for us below, 2Al (s) + 3I2 (s) → 2AlI3 (s) We know from the equation that 2 moles of aluminum requires 3 moles of I2. This can be written as the mole ratio, 3 moles I2 / 2 moles Al The problem states that we have 35g of aluminum. We must convert this to moles using the average atomic mass of aluminum, 35.0g Al x (1 mol Al/ 26.98g Al) = 1.30 moles of Al Now that we have the moles of Al required, we can find the moles of I2 required, 1.30 moles Al x (3 moles I2/ 2 moles Al) = 1.95 moles I2 Now, we must convert this mole value to mass using I2’s atomic mass. Atomic mass of I= 126.9g/mol Atomic mass of I2= 253.8g/mol 1.95 moles I2 x (253.8g/mol) = 495g of I2 Therefore, we need 495g of I2 to react with 35g of Al.

9-5: Calculations Involving a Limiting Reactant Example Question Consider the production of hydrogen for use in the manufacturing of ammonia. Ammonia is made by combining nitrogen from the air with hydrogen. The hydrogen for this process is produced by the reaction of methane with water according to the equation: CH4 (g) + H2O (g) → 3H2 (g) + CO (g) What mass of water is required to react exactly with 249g of methane? We must first convert the mass of methane to moles, 249g CH4 x (1 mol/ 16.04g CH4) = 15.5 moles CH4 We can see from the balanced equation that 1 mole of CH4 reacts with 1 mole of H2O. Because of this, we would have 15.5 moles of H2O per 15.5 moles of CH4. Now we must convert 15.5 moles of H2O to grams. Molar mass of H2O= 18.016g/mol 15.5 moles H2O x (18.016g/mol) = 279g of H2O Therefore, 279g of H2O is required to react exactly with 249g of CH4, meaning both reactants will “run out” at the same time. If, on the other hand, 249g of methane is mixed with 300g of H2O, the methane will be consumed before the water runs out, and the water would be in excess. We say here that methane is the limiting reactant, as it limits the amount of product that can be formed.

Limiting Reactant: The reactant that runs out first, thus limiting the amount of product that can be formed.

Example Question 9.8 Question: Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. The other products in this reaction are solid copper and water vapour. How many grams of N2 are formed when 18.1g of NH3 is reacted with 90.4g of CuO? Solution: We must first start by writing the balanced equation for the reaction. We can start this by writing out the chemical formulas of all the reactants and products, and then balancing. 2NH3 (g) + 3CuO (s) → N2 (g) + 3Cu (s) + 3H2O (g) Now, we must convert the masses of the reactants to moles. 18.1g NH3 x (1 mole/17.03g) = 1.06 moles of NH3 90.4g CuO x (1 mole/79.55g) = 1.14 moles of CuO Now, we must determine which reactant is limiting. We do this by using the mole ratio of CuO to NH3. 1.06 moles NH3 x (3 moles CuO/ 2 moles NH3) = 1.59 moles of CuO needed Now, we compare the amount of CuO needed to the amount of CuO available. There are less moles of CuO available (1.14 moles) than is needed (1.59 moles) to fully react with the NH3. This means that the CuO is the limiting reactant. Because of this, we must use the CuO mole value with the mole ratio between CuO and N2 to calculate the moles of N2 formed. 1.14 moles CuO x (1 mol N2/3 moles CuO) = 0.380 moles of N2 Now, we must convert the moles of N2 formed to mass. 0.380 moles N2 x (28.02g/mol) = 10.6g N2 Therefore, 10.6g of N2 can be produced with 18.1g NH3 and 90.4g CuO.

9-6: Percent Yield The amount of product formed when the limiting reactant runs out is called the theoretical yield of that product. However, the amount of product predicted is seldom ever obtained. One reason for this is the presence of side reactions. The actual yield of the product is the amount of product actually obtained. This is often compared to the theoretical yield. This comparison, expressed as a percent, is called percent yield. The equation for calculating percent yield is included below, Percent Yield= (Actual Yield / Theoretical Yield) x 100%...