Let's try a simple example. Reactant A is a test tube. I have 20 of them. Reactant B is a stopper. I have 30 of them. Product C is a stoppered test tube. The reaction is: A + B ---> C or: test tube plus stopper gives stoppered test tube. So now we let them &quot;react.&quot; The first stopper goes in, the second goes in and so on. Step by step we use up stoppers and test tubes (the amounts go down) and make stoppered test tubes (the amount goes up). Suddently, we run out of one of the &quot;reactants.&quot; Which one? That's right. We run out of test tubes first. Seems obvious, doesn't it? We had 20 test tubes, but we had 30 stoppers. So when the test tubes are used up, we have 10 stoppers sitting thereunused. And we also have 20 test tubes with stoppers firmly inserted. So,what &quot;reactant&quot; is limiting and which is in excess?
If we have 4 slices of bread and 2 slices of cheese, we can make exactly 2 sandwiches with nothing left. This has no limiting reactant. However, if we have 6 slices of bread and 2 slices of cheese, we can only make 2 sandwiches. We have enough bread to make 3 sandwiches but only enough cheese to make 2 sandwiches. The amount of cheese limits the amount of sandwiches we can make. The cheese is the limiting reactant and the bread is the excess reactant. In another example, if we have 8 slices of bread and 5 slices of cheese, we have enough bread to make 4 sandwiches and enough cheese to make 5 sandwiches. The bread is the limiting reactant in this example and cheese is the excess reactant, since some cheese will be left over.
1. Write a balanced equation for the reaction. 2 H2 + O2 2 H2O 2. Convert both reactant quantities to moles. 3. Using the mole ratio from the equation, determine the moles of water that could be formed by each reactant. 4. Oxygen produces the least amount of water. 16 grams of oxygen cannot produce as much water as 4 grams of hydrogen. In other words, 16 grams of oxygen will be used up in the reaction before 4 grams of hydrogen. Oxygen is the &quot;limiting&quot; reactant. Use oxygen for the calculation of product amount. 5. Complete the problem by converting moles of H2O to mass of H2O.
1. Write a balanced equation for the reaction. 2 H2 + O2 2 H2O 2. Convert both reactant quantities to moles. 3. Using the mole ratio from the equation, determine the moles of water that could be formed by each reactant. 4. Oxygen produces the least amount of water. 16 grams of oxygen cannot produce as much water as 4 grams of hydrogen. In other words, 16 grams of oxygen will be used up in the reaction before 4 grams of hydrogen. Oxygen is the &quot;limiting&quot; reactant. Use oxygen for the calculation of product amount. 5. Complete the problem by converting moles of H2O to mass of H2O.
1. Write a balanced equation for the reaction. 2 H2 + O2 2 H2O 2. Convert both reactant quantities to moles. 3. Using the mole ratio from the equation, determine the moles of water that could be formed by each reactant. 4. Oxygen produces the least amount of water. 16 grams of oxygen cannot produce as much water as 4 grams of hydrogen. In other words, 16 grams of oxygen will be used up in the reaction before 4 grams of hydrogen. Oxygen is the &quot;limiting&quot; reactant. Use oxygen for the calculation of product amount. 5. Complete the problem by converting moles of H2O to mass of H2O.
1. Write a balanced equation for the reaction. 2 H2 + O2 2 H2O 2. Convert both reactant quantities to moles. 3. Using the mole ratio from the equation, determine the moles of water that could be formed by each reactant. 4. Oxygen produces the least amount of water. 16 grams of oxygen cannot produce as much water as 4 grams of hydrogen. In other words, 16 grams of oxygen will be used up in the reaction before 4 grams of hydrogen. Oxygen is the &quot;limiting&quot; reactant. Use oxygen for the calculation of product amount. 5. Complete the problem by converting moles of H2O to mass of H2O.
1. Write a balanced equation for the reaction. 2 H2 + O2 2 H2O 2. Convert both reactant quantities to moles. 3. Using the mole ratio from the equation, determine the moles of water that could be formed by each reactant. 4. Oxygen produces the least amount of water. 16 grams of oxygen cannot produce as much water as 4 grams of hydrogen. In other words, 16 grams of oxygen will be used up in the reaction before 4 grams of hydrogen. Oxygen is the &quot;limiting&quot; reactant. Use oxygen for the calculation of product amount. 5. Complete the problem by converting moles of H2O to mass of H2O.
1 - calcium oxide 2. 0.78g of aluminum oxide 3a - AgNO3 3b - 26.6 g AgNO3 4a - H3PO4 4b - 2.85 g H2O 4c - 0.106 mole Na3PO4 4d - 10.15 g SO2
Stoichiometry
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Chemistry 30S Unit 3 – Chemical Reactions
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<ul><li>Recall that the coefficients in a balanced chemical equation show us in what proportions the reactants combine and products are made. Let’s look at the first equation we balanced: </li></ul>CHEMICAL REACTIONS Balanced Chemical Equations Review 2 H 2 ( g ) + O 2 ( g ) 2 H 2 O ( g )
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<ul><li>We will consider the coefficients to be number of moles that combine or are produced. Thus they are also called molar coefficients . If we have moles, we can now convert moles to mass and volume at STP: </li></ul>CHEMICAL REACTIONS Molar Coefficients
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<ul><li>Stoichiometry is the use of proportions or ratios to determine quantities, usually masses and volumes, of reactants and products in a chemical reaction. </li></ul><ul><li>Stoichiometry comes from the Greek words stoicheion (meaning "element") and metron (meaning "measure"). We’ve seen the root metron in words like thermometer, calorimeter, micrometer, etc. These are instruments used to make measurements. </li></ul><ul><li>From the Greek, stoichiometry means measuring the elements of a reaction. </li></ul>CHEMICAL REACTIONS Defining Stoichiometry
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<ul><li>Substances combine and are produced in distinct whole number ratios. We make these ratios using the coefficients of the balanced chemical equation. Recall, in our example of the synthesis of water, </li></ul><ul><li>2 H 2 ( g ) + O 2 ( g ) 2 H 2 O( g ) </li></ul><ul><li>The ratio of the coefficients in a chemical equation is known as the molar ratio for those two substances. </li></ul>CHEMICAL REACTIONS The Molar Ratio
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<ul><li>We use stoichiometry to determine amounts of reactants that combine and products that are formed in a chemical reaction. </li></ul><ul><li>The molar ratio for the substances involved helps us determine quantities, in terms of moles. We can determine how many moles of a reactant will react with a given amount of another reactant or the moles of product produced. When formulating the molar ratio always go by </li></ul>CHEMICAL REACTIONS Solving Stoichiometry Problems
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<ul><li>To solve stoichiometry problems involving moles: </li></ul><ul><ul><li>Step 1: Balance the chemical equation, if not already done. </li></ul></ul><ul><ul><li>Step 2: Determine the molar ratio for the substances involved. </li></ul></ul><ul><ul><li>Step 3: Multiply the given amount by the molar ratio. </li></ul></ul>CHEMICAL REACTIONS Solving Stoichiometry Problems
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<ul><li>Example Problem #1 </li></ul><ul><li>For the following equation, </li></ul><ul><li>H 2 ( g ) + O 2 ( g ) H 2 O( g ) </li></ul><ul><ul><li>How many moles of oxygen gas will react with 5.0 moles of hydrogen gas? </li></ul></ul><ul><ul><li>How many moles of water will be produced from 7.0 moles of oxygen gas? </li></ul></ul>CHEMICAL REACTIONS Solving Stoichiometry Problems
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<ul><li>The coefficients in a balanced chemical equation are the ratios of MOLES that are reacting and produced. We cannot use masses or volumes of reactants or products with the molar ratio. It is important that any quantity be converted to moles before using the molar ratio. When solving stoichiometry problems follow these steps: </li></ul>CHEMICAL REACTIONS Stoichiometry – Part 2
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<ul><li>Step 1: If not already balanced, balance the chemical equation. </li></ul><ul><li>Step 2: If necessary, convert the given amount from mass or volume to moles. </li></ul><ul><li>Step 3: Use the molar ratio to calculate the amount of substance requested. </li></ul><ul><li>Step 4: If needed, convert the moles of requested substance to mass or volume. </li></ul>CHEMICAL REACTIONS Stoichiometry – Part 2
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<ul><li>Example Problem #1 </li></ul><ul><li>What mass of NaCl is formed from the reaction between 3.20 moles of chlorine gas and an excess of sodium, according to the equation below? </li></ul><ul><li>Na( s ) + Cl 2 ( g ) NaCl( s ) </li></ul>CHEMICAL REACTIONS Solving Mass – Mol Stoichiometry Problems
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<ul><li>Example Problem #2 </li></ul><ul><li>Calculate the moles of hydrogen needed to make 15.0 g of ammonia? </li></ul><ul><li>N 2 ( g ) + 3 H 2 ( g ) 2 NH 3 ( g ) </li></ul>CHEMICAL REACTIONS Solving Mass – Mol Stoichiometry Problems
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<ul><li>Example Problem #1 </li></ul><ul><li>What mass of carbon dioxide is produced from burning 50.0 g of ethane, C 2 H 6 ? </li></ul><ul><li>C 2 H 6 ( g ) + O 2 ( g ) CO 2 ( g ) + H 2 O( g ) </li></ul>CHEMICAL REACTIONS Solving Mass – Mass Stoichiometry Problems
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<ul><li>Example Problem #2 </li></ul><ul><li>Given the equation </li></ul><ul><li>FeS( s ) + O 2 ( g ) Fe 2 O 3 ( s ) + SO 2 ( g ) </li></ul><ul><li>How many grams of FeS are needed to make 36.3 g of Fe 2 O 3 ? </li></ul>CHEMICAL REACTIONS Solving Mass – Mass Stoichiometry Problems J k
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<ul><li>Example Problem #1 </li></ul><ul><li>Given the equation </li></ul><ul><li>4 FeS( s ) + 7 O 2 ( g ) 2 Fe 2 O 3 ( s ) + 4 SO 2 ( g ) </li></ul><ul><li>What volume of O 2 , in Litres, is needed to completely react with 25.0 g of FeS, at STP? </li></ul>CHEMICAL REACTIONS Solving Volume – Mass Stoichiometry Problems H
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<ul><li>Example Problem #2 </li></ul><ul><li>According to the equation below, what mass of ammonia can be produced from 50.0 L of nitrogen gas, at STP. </li></ul><ul><li>N 2 ( g ) + 3 H 2 ( g ) 2 NH 3 ( g ) </li></ul>CHEMICAL REACTIONS Solving Volume – Mass Stoichiometry Problems H
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What is the Limiting Reagent? <ul><li>It is simply the substance in a chemical reaction that runs out first. </li></ul><ul><li>The limiting reagent is the reactant that is completely consumed. </li></ul><ul><li>It is called the limiting reactant because when it is all gone no more product can be created – the reactant limits the amount of product formed. </li></ul>
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Limiting Reagents An analogy of limiting reactants could be the making of a cheese sandwich. Making this cheese sandwich (B 2 C) requires 2 slices of bread (B) and 1 slice of cheese (C). The equation for this sandwich would be 2 slices of bread + 1 slice of cheese = 1 cheese sandwich or 2 B + 1 C 1 B 2 C
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Limiting Reagents 2 C 8 H 18 (l) + 25 O 2 (g) 16 CO 2 (g) + 18 H 2 O(l)
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Steps in solving a limiting reactant problem <ul><li>Step 1:If necessary, find the number of moles of each reactant. </li></ul><ul><li>Step 2:Using the molar coefficients, determine how much of one reactant is needed to use up the other. </li></ul><ul><li>Step 3:If the amount in step 2 is more than what is given of that reactant, this reactant is the limiting reactant. </li></ul><ul><li>Step 4:Use the molar coefficients and the given amount of the limiting reactant to determine the moles of product. </li></ul><ul><li>Step 5:Convert moles of product to mass or volume. </li></ul>
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Limiting Reagent Problems <ul><li>What mass of water can be produced by 4g of H 2 reacting with 16g of O 2 ? </li></ul>H J
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Limiting Reagent Problems <ul><li>2H 2 + O 2 2H 2 0 </li></ul><ul><li>Step 1 : Find the Number of moles of each </li></ul><ul><li>Step 2 : Using Molar Coefficients, determine how much of one reactant is needed to use up the other. </li></ul><ul><li>Step 3 : If the amount in step 2 is more than what is given of that reactant, this reactant is the limiting reactant. </li></ul>h h h
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Limiting Reagent Problems <ul><li>2H 2 + O 2 2H 2 0 </li></ul><ul><li>Step 4 : Use the limiting reactant and the molar coefficient to determine the number of moles of product. </li></ul><ul><li>Step 5 : Convert the moles of product to mass. </li></ul>Hg
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Work these limiting reactant practice problems: <ul><li>1. Seventy five grams of calcium oxide react with one hundred thirty grams of hydrochloric acid to produce a salt and water. What is the limiting reactant? </li></ul><ul><li>2. How much aluminum oxide are produced when 46.5g of Al react with 165.37g of MnO? </li></ul><ul><li>3. Five grams of copper metal react with a solution containing twenty grams of silver nitrate to produce copper (II) nitrate and silver. </li></ul><ul><li>a.What is the limiting reactant? </li></ul><ul><li>b.How much of the limiting reactant would be needed to react completely with the given amount of excess reactant? </li></ul><ul><li>4. A solution containing 20.0 g of sodium sulfite reacts with 7.0 ml of phosphoric acid. The concentration of the acid solution is such that there are 1.83 grams of H3PO4 per milliliter of solution. Determine the following: </li></ul><ul><li>a.The mass of the excess reactant remaining at completion. </li></ul><ul><li>b.Grams of water produced. </li></ul><ul><li>c.Moles of sodium phosphate produced. </li></ul><ul><li>d.Grams of sulfur dioxide produced. </li></ul>