Conservation of Matter Lesson 7-8 PDF

LESSON 7-8 Conservation of Matter **INTRODUCTION** Chemical equation is indeed useful in interpreting a chemical reaction. Now for this particular lesson, you will be focusing to the reactant and the product that are involved in a chemical reaction. In addition, you will also understand how this...

LESSON 7-8 Conservation of Matter **INTRODUCTION** Chemical equation is indeed useful in interpreting a chemical reaction. Now for this particular lesson, you will be focusing to the reactant and the product that are involved in a chemical reaction. In addition, you will also understand how this can be used to calculate the percent yield and theoretical yield and how you can properly identify the limiting reagent and the excess reagent. **LESSON PROPER** **What is Conservation of Matter?** **In our previous lesson, you already encountered the law of conservation of mass or also known as the law of conservation of matter. It states that the amount of matter before and after a reaction is always equal.** **Calculation that is involved in a Balanced Equation:** **a.) Mole-to-Mole Conversion** **The number of moles of any substance in the chemical reaction or chemical equation can be determined if the number of moles of one substance is given in the problem. This is possible using the mole conversion factor based on the coefficients in the balanced chemical equation.** **Step in Mole-to-Mole Conversion** **Example:** [**4NH**~**3**~]{.math.inline} **+** [**3O**~**2**~]{.math.inline} **=** [**2N**~**2**~]{.math.inline} **+** [**6H**~**2**~**O**]{.math.inline} **How many moles of** [*O*~2~]{.math.inline} **are needed to react with 8 moles of** [NH~3~]{.math.inline}**?** **Step 1: Analyze the given chemical equation.** --------------------------------------------------- **Substance** **Number of Mole** ------------------------------ -------------------- \ **4** [NH~3~]{.math.display}\ \ **3** [*O*~2~]{.math.display}\ \ **2** [*N*~2~]{.math.display}\ \ **6** [*H*~2~*O*]{.math.display}\ --------------------------------------------------- **Step 2: Identify the problem.** **How many moles of** [*O*~2~]{.math.inline} **are needed to react with 8 moles of** [NH~3~]{.math.inline}**?** **\_\_\_\_ moles of** [*O*~2~]{.math.inline} **: 8 moles of** [NH~3~]{.math.inline} **Step 3: Identify the conversion factor based from the balanced chemical equation** [4*NH*~3~]{.math.inline} **+** [3*O*~2~]{.math.inline} **=** [2*N*~2~]{.math.inline} **+** [6*H*~2~*O*]{.math.inline} **Based on the chemical equation for every 3 moles of** [*O*~2~]{.math.inline} **there will be 4 moles of** [NH~3~]{.math.inline}**.** [\$\\frac{{3\\ mole\\ O}\_{2}}{{4\\ mole\\ NH}\_{3}}\$]{.math.inline} **or** [\$\\frac{{4\\ mole\\ NH}\_{3}}{{3\\ mole\\ O}\_{2}}\$]{.math.inline} **Step 4: Multiply to the unit analysis to get the conversion factor.** \ [\$\${8\\ mole\\ NH}\_{3}\\ \\ \\ \\times \\ \\ \\ \\ \\ \\frac{{3\\ mole\\ O}\_{2}}{{4\\ mole\\ NH}\_{3}}\\text{\\ \\ }\$\$]{.math.display}\ **Note: Choose the unit analysis where you can cancel out the given mole in the problem.** [\${8\\ mole\\ NH}\_{3}\\ \\ \\ \\times \\ \\ \\ \\ \\ \\frac{{3\\ mole\\ O}\_{2}}{{4\\ mole\\ NH}\_{3}}\$]{.math.inline} **=** [**6** **mole** **O**~**2**~]{.math.inline} ***Question:* How many moles of** [*O*~2~]{.math.inline} **are needed to react with 8 moles of** [NH~3~]{.math.inline}**?** - ***It needed 6 mole of*** [*O*~2~]{.math.inline} ***to react with 8 moles of***[ NH~3~]{.math.inline}***.*** **b.) Mole-to-Mass Conversion** ***Grams* - is the most convenient unit for measuring the amount of a reactant or products in a laboratory. In order to convert moles to grams or grams to moles, you have to get or use the molar mass of the given compound.** **Step in Mole-to-Mass Conversion** **Example:** [**4NH**~**3**~]{.math.inline} **+** [**3O**~**2**~]{.math.inline} **=** [**2N**~**2**~]{.math.inline} **+** [**6H**~**2**~**O**]{.math.inline} **How many grams of** [*O*~2~]{.math.inline} **are needed to react with 8 moles of** [NH~3~]{.math.inline}**?** **Step 1: Identify the problem.** **How many grams of** [*O*~2~]{.math.inline} **are needed to react with 8 moles of** [NH~3~]{.math.inline}**?** **\_\_\_\_ moles of** [*O*~2~]{.math.inline} **: 8 moles of** [NH~3~]{.math.inline} **Step 2: Identify the conversion factor based from the balanced chemical equation** [4*NH*~3~]{.math.inline} **+** [3*O*~2~]{.math.inline} **=** [2*N*~2~]{.math.inline} **+** [6*H*~2~*O*]{.math.inline} **Based on the chemical equation for every 3 moles of** [*O*~2~]{.math.inline} **there will be 4 moles of** [NH~3~]{.math.inline}**.** [\$\\frac{{3\\ mole\\ O}\_{2}}{{4\\ mole\\ NH}\_{3}}\$]{.math.inline} **or** [\$\\frac{{4\\ mole\\ NH}\_{3}}{{3\\ mole\\ O}\_{2}}\$]{.math.inline} **Step 3: Find out the number of moles of**[ *O*~2~]{.math.inline} **needed to react to 8 moles of** [NH~3~]{.math.inline} **using the unit analysis.** [\${8\\ mole\\ NH}\_{3}\\ \\ \\ \\times \\ \\ \\ \\ \\ \\frac{{3\\ mole\\ O}\_{2}}{{4\\ mole\\ NH}\_{3}}\$]{.math.inline} **=** [**6** **mole** **O**~**2**~]{.math.inline} **Step 4: Get the molar mass of** [*O*~2~]{.math.inline}**.** **Element** **Average Atomic Mass (or Average Molecular Mass) (*u*)** **Unified Mass Unit (*u*)** **Molar Mass Constant (**[**M**~**u**~]{.math.inline}**)** **Molar Mass (1 g mol^−1^)** ------------- ----------------------------------------------------------- --- ----------------------------- --- ------------------------------------------------------------- --- -------------------------------------------------- Oxygen (O) 15.9 u ÷ 1 u × 1 g mol^−1^ = 15.9 [\$\\frac{g}{\\text{mol}}\$]{.math.inline} **Since there is two (2) oxygen atom it will be multiplied into two (2).** **Atom** **Number of Atom** **Molar Mass** **Molar mass of** [**O**~**2**~]{.math.inline} ---------- -------------------- ------- -------------------------------------------------- ------- ------------------------------------------------------ O **2** **˟** 15.9 [\$\\frac{g}{\\text{mol}}\$]{.math.inline} **=** **31.8** [\$\\frac{g}{\\text{mol}}\$]{.math.inline} **Step 5: Change the moles of** [*O*~2~]{.math.inline} **to grams using the molar mass of** [*O*~2~]{.math.inline} **as conversion factor.** [\${6\\ mole\\ O}\_{2}\\ \\ \\ \\times \\ \\ \\ \\ \\ \\frac{31.8\\ \\text{\~g\~O}\_{2}}{{1\\ mole\\ O}\_{2}}\$]{.math.inline} **=** [**190.8** **g** **O**~**2**~]{.math.inline} ***Question:* How many grams of** [*O*~2~]{.math.inline} **are needed to react with 8 moles of** [NH~3~]{.math.inline}**?** - ***It needs 190.8 grams of*** [*O*~2~]{.math.inline} ***to react with 8 moles of*** [NH~3~.]{.math.inline} **c.) Mass-to-Mass Conversion** **This conversion happens if the problem asks for mass in grams of one substance and the mass of another substance is given.** **Step in Mass-to-Mass Conversion** **Example:** [**4NH**~**3**~]{.math.inline} **+** [**3O**~**2**~]{.math.inline} **=** [**2N**~**2**~]{.math.inline} **+** [**6H**~**2**~**O**]{.math.inline} **How many g of** [*O*~2~]{.math.inline} **are needed to react with 250 g of** [NH~3~]{.math.inline}**?** **Step 1: Identify the problem.** **How many grams of** [*O*~2~]{.math.inline} **are needed to react with 250 grams of** [NH~3~]{.math.inline}**?** **\_\_\_\_ g of** [*O*~2~]{.math.inline} **: 250 g of** [NH~3~]{.math.inline} **Step 2: Get the molar mass of** [NH~3~]{.math.inline}**.** **Atom** **Number of Atom** **Molar Mass** ---------- -------------------- ------- ---------------------------------------------- ------- ---------- N **1** **˟** 14.0  **=** **14.0** H **3** **˟** 1.0  **=** **3.0** **Molar mass of** [NH~**3**~]{.math.inline} **=** **17.0** **Step 3: Change the grams of** [NH~3~]{.math.inline} **to moles of** [NH~3~]{.math.inline}**.** [\${250\\ g\\ NH}\_{3}\\ \\ \\ \\times \\ \\ \\ \\ \\ \\frac{1\\ \\text{mole}\\text{\~NH}\_{3}}{17.0\\ g\\text{\~NH}\_{3}}\$]{.math.inline} **=** [14.7 *mole* *NH*~3~ ]{.math.inline} **Step 4: Change the mole of** [NH~3~]{.math.inline} **to moles of** [*O*~2~]{.math.inline}**.** [\${14.7\\ mole\\ NH}\_{3}\\ \\ \\ \\times \\ \\ \\ \\ \\ \\frac{3\\ \\text{mole}\\text{\~O}\_{2}}{4\\ mole\\text{\~NH}\_{3}}\$]{.math.inline} **=** [11.0 *mole* *O*~2~ ]{.math.inline} **Note: Review the chemical equation to get the conversion factor.** [**4NH**~**3**~]{.math.inline} **+** [**3O**~**2**~]{.math.inline} **=** [**2N**~**2**~]{.math.inline} **+** [**6H**~**2**~**O**]{.math.inline} **Step 5: Get the molar mass of** [*O*~2~]{.math.inline}**.** **Atom** **Number of Atom** **Molar Mass** **Molar mass of** [**O**~**2**~]{.math.inline} ---------- -------------------- ------- -------------------------------------------------- ------- ------------------------------------------------------ O **2** **˟** 15.9 [\$\\frac{g}{\\text{mol}}\$]{.math.inline} **=** **31.8** [\$\\frac{g}{\\text{mol}}\$]{.math.inline} **Step 6: Change the moles of** [*O*~2~]{.math.inline} **to grams of** [*O*~2~]{.math.inline}**.** [\${11.0\\ mole\\ O}\_{2}\\ \\ \\ \\times \\ \\ \\ \\ \\ \\frac{31.8\\ \\ g\\text{\~O}\_{2}}{1\\ mole\\text{\~O}\_{2}}\$]{.math.inline} **=** [**42.8** **g** **O**~**2**~ ]{.math.inline} ***Question:* How many g of** [*O*~2~]{.math.inline} **are needed to react with 250 g of** [NH~3~]{.math.inline}**?** - ***It needs***[ ]{.math.inline} ***42.8 grams of*** [*O*~2~.]{.math.inline} **During a chemical reaction, it is set to produce the maximum quantity of useful products from the original substance or what we call the *reactants* at a minimum cost. Reactants could be a limiting reactant or an excess reactant.** - **Limiting Reactant -- is the reactant that limits the reaction** - **Excess Reactant -- is the reactant that contains one or more needed to react with limiting reactant.** - **Product - is a substance that formed during the chemical reaction** **Reaction Yield** - ***Theoretical Yield* is the maximum amount of the product that can be produced from the given amount of reactant.** - ***Experimental or Actual Yield* is the amount of product obtained in a laboratory from a reaction. Due to various factors like impurities of the substances, incomplete reactions and other side reaction these may cause the loss of some products.** - ***Percentage Yield* expressed the efficiency of a reaction. These gives an idea if a product is a good investment or not.** \ [\$\$Percent\\ yield = \\frac{\\text{actual\\ yield}}{\\text{theoretical\\ yield}}x100\$\$]{.math.display}\ **For example: JC was to conduct an experiment, supposed that he has 3.5 grams of aluminum and 4.5 grams of sodium chloride. Find the following:** Al + NaCl [AlCl~3~]{.math.inline} + Na **a. Which is the limiting reactant?** **b. How much excess reactant remains after the reaction?** **c. How many grams of** [AlCl~3~]{.math.inline} are formed? d\. What will be the percentage yield if the actual yield is 2.5 grams of [AlCl~3~]{.math.inline}? Solution: **Step 1**: Balanced the chemical equation. Al + 3NaCl [AlCl~3~]{.math.inline} + 3Na **Step 2**: Calculate the molar mass of the reactants. Al = 27 [\$\\frac{g}{\\text{mol}}\$]{.math.inline} ------ --- ---------------------------------------------------------- -- NaCl 35 [\$\\frac{g}{\\text{mol}}\$]{.math.inline} 23 [\$\\frac{g}{\\text{mol}}\$]{.math.inline} 58 [\$\\frac{g}{\\text{mol}}\$]{.math.inline} *of NaCl* **Step 3: Divide the calculated molar mass of the given element or compound from the given mass. And then, the number of moles are divided by the coefficient indicated in the balanced chemical equation.** **Al =** [\$\\frac{3.5\\ g}{27\\ g/mol}\$]{.math.inline} **= 0.13 mol** **Al =** [\$\\frac{0.13\\ mol}{1}\$]{.math.inline} **= 0.13 mol** ------------------------------------------------------------- ---------------- ------------------------------------------------------ ---------------- **NaCl =** [\$\\frac{4.5\\ g}{58\\ g/mol}\$]{.math.inline} **= 0.08 mol** **NaCl =**[\$\\frac{0.08\\ mol}{3}\$]{.math.inline} **= 0.03 mol** ***a. Which is the limiting reactant?*** - **The limiting reactant is NaCl since it has a lesser calculated value that Al.** ***b. How much excess reactant remains after the reaction?*** ----------------------------------------- **0.13 mol - 0.03 mol = 0.1 mol of Al** ----------------------------------------- - **There are 0.1 mol left in the excess reactant which is the Al.** ***c. How many grams of*** [AlCl~3~]{.math.inline} *are formed?* **Get the molar mass of** [AlCl~3~]{.math.inline}. --------------------------------------------------------------------------------------------------------------------------------------- \ [AlCl~3~]{.math.display}\ ---------------------------- ----- ------------------------------------------------ --------------------------------------------------- 3 ˟ 35 [\$\\frac{g}{\\text{mol}}\$]{.math.inline} = 105 [\$\\frac{g}{\\text{mol}}\$]{.math.inline} 1 ˟ 27 [\$\\frac{g}{\\text{mol}}\$]{.math.inline} = 27 [\$\\frac{g}{\\text{mol}}\$]{.math.inline} 132 [\$\\frac{g}{\\text{mol}}\$]{.math.inline} --------------------------------------------------------------------------------------------------------------------------------------- ------------------------------- ------------------ ----------------------------------------------------------------- ------------------------------------------------------------------------------- -------------------------------------------------------------------------------------------------- -------------------------------------- g of [AlCl~3~]{.math.inline} **= 4.5 g NaCl** ˟ [\$\\frac{1mol\\ \\ NaCl}{58\\ g\\ \\ NaCl}\$]{.math.inline} ˟ [\$\\frac{1mol\\ \\ \\text{AlCl}\_{3}}{3\\ mol\\ \\ NaCl}\$]{.math.inline} ˟ [\$\\frac{132\\ g\\ \\ \\text{AlCl}\_{3}}{\\ 1\\ mol\\ \\ \\text{AlCl}\_{3}}\$]{.math.inline} = 3.41 g of [AlCl~3~]{.math.inline} ------------------------------- ------------------ ----------------------------------------------------------------- ------------------------------------------------------------------------------- -------------------------------------------------------------------------------------------------- -------------------------------------- - **There are 3.41 grams of** [AlCl~3~]{.math.inline} formed in the reaction. *d. What will be the percentage yield if the actual yield is 2.45 grams of* [AlCl~3~]{.math.inline}*?* [\$Percentage\\ yield = \\frac{\\text{actual\\ yield}}{\\text{theoretical\\ yield}}x100\$]{.math.inline} [\$Percentage\\ yield = \\frac{2.45\\ g}{3.41\\ g}x100\$]{.math.inline} [*Percentage* *yield* = 78 %]{.math.inline} - **The percentage yield is 78 %**.