# Starter S-125 1.2 moles NaC 2 H 3 O 2 are used in a reaction. How many grams is that?

## Presentation on theme: "Starter S-125 1.2 moles NaC 2 H 3 O 2 are used in a reaction. How many grams is that?"— Presentation transcript:

Starter S-125 1.2 moles NaC 2 H 3 O 2 are used in a reaction. How many grams is that?

Stoichiometry Chapter 12

12.1 The Arithmetic of Equations Chapter 12

The basis for solving stoichiometry problems is a balanced chemical reaction A balanced reaction is used to calculate How much reactant is needed How much product is produced 12.1 The Arithmetic of Equations

Stoichiometry – the calculation of quantities This reaction produces Ammonia which is used in fertilizers Balanced reactions are usually used to calculate grams of product or reactant 12.1 The Arithmetic of Equations

Analysis of the reaction Atoms – 2 atoms of nitrogen combine with 6 atoms of hydrogen – product is 2 nitrogen and 6 hydrogen Molecules – 1 molecule of nitrogen gas combines with 3 molecules of hydrogen gas to produce 2 molecules of ammonia 12.1 The Arithmetic of Equations

Analysis of the reaction Most useful Moles – 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia 12.1 The Arithmetic of Equations

Analysis of the reaction Mass is always conserved in a chemical reaction 12.1 The Arithmetic of Equations

Analysis of the reaction The key moles can be converted to grams grams can be converted to moles Volume – remember that one mole of gas at STP is 22.4L 12.1 The Arithmetic of Equations

Analysis of the reaction The key moles can be converted to grams grams can be converted to moles Volume – remember that one mole of gas at STP is 22.4L 12.1 The Arithmetic of Equations

Analysis of the reaction So if we start with 50g of N 2, how many moles do we have How many moles of NH 3 would be produced? For every 1 mole of N 2, 2 moles of NH 3 12.1 The Arithmetic of Equations

Analysis of the reaction How many grams of ammonia are produced 12.1 The Arithmetic of Equations

Starter S-126 A. Balance the following reactions H 2 SO 4 + KOH  K 2 SO 4 + H 2 O B.What is the molar mass of the sulfuric acid. C.If 20.0g of sulfuric acid is used, how many moles is that? D.How many moles of water would be produced?

12.2 Chemical Calculations Chapter 12

12.2 Chemical Calculations Mole Ratios Come from balanced chemical reactions Conversion factors derived from the coefficients in the balanced reactions 12.2 Chemical Calculations

Mole-Mole Calculations It is possible to convert from one quantity in a balanced reaction to another using mole ratios For example – if 3.7 moles of sulfur dioxide is produced, how many moles of oxygen were used? 12.2 Chemical Calculations

Mass-Mass Calculations Three steps 1.Convert given mass values into mole values. 2.Use a mole ratio to convert to the moles that the question requests 3.Convert this mole quantity to a mass value 12.2 Chemical Calculations

Mass-Mass Calculations How many grams of Oxygen are needed to produce 30.0g of Sulfur Dioxide? 12.2 Chemical Calculations

Volume-Mass Calculations Same steps, but volume is converted to moles, or moles to volume Example: If 4.0L of nitrogen monoxide reacts, how many grams of oxygen are used? 12.2 Chemical Calculations

Starter S-127 A. Balance the following reactions Al(NO 3 ) 3 + Na 2 SO 4  Al 2 (SO 4 ) 3 +NaNO 3 B. If 50.0g of Aluminum Nitrate reacts, how many grams of Aluminum Sulfate are produced?

12.3 Limiting Reagent and Percent Yield Chapter 12

Limiting Reagent – the reactant that runs out first Amounts of both reactants are given Example: 80.0g Copper, 25.0g Sulfur 1.Calculate how many moles would each reactant produce Copper 12.3 Limiting Reagent and Percent Yield

Limiting Reagent – the reactant that runs out first Amounts of both reactants are given Example: 80.0g Copper, 25.0g Sulfur 1.Calculate how many moles would each reactant produce Sulfur 12.3 Limiting Reagent and Percent Yield

Limiting Reagent – the reactant that runs out first From the reactions Copper would produce 0.630 mol Cu 2 S Sulfur would produce 0.780 mol Cu 2 S That means copper will run out first – it is the limiting reagent Sulfur would be the excess reagent 12.3 Limiting Reagent and Percent Yield

One for you now 1.Balance the reaction 2.If 75.6g C 2 H 4 reacts with 100.8g O 2 – what is the limiting reagent? Oxygen 3.How many moles of water are produced? 2.10 mol H 2 O 4.How many grams of water are produced? 12.3 Limiting Reagent and Percent Yield

Starter S-132 A. Balance the following reactions SiO 2 + C  SiC + CO B. 35 grams of silicon dioxide reacts with 10.0g of Carbon, how much carbon monoxide is formed?

Reactions rarely produce as much product as is predicted -reactants can be impure -reactions may not go to completion -may compete with smaller “side” reactions In some reactions as little as 60% yield is considered a good result 12.3 Limiting Reagent and Percent Yield

Yield – how much product is produced Theoretical Yield – the value calculated using stoichiometry Actual Yield – the amount of product that actually forms 12.3 Limiting Reagent and Percent Yield

Percent Yield – a ratio of actual to theoretical yield This number must be 100% or less In lab the actual yield is the result you get On a test, it will be a number that is given to you The theoretical yield is calculated using limiting reactants 12.3 Limiting Reagent and Percent Yield

Example: What is the theoretical yield of Calcium Oxide if 24.8g of Calcium Carbonate decomposes in the following reaction. Balanced 12.3 Limiting Reagent and Percent Yield

Example: What is the percent yield if actual yield is 9.6g? 12.3 Limiting Reagent and Percent Yield

Example: What is the theoretical yield if 15.0g of nitrogen reacts with 15.0g of hydrogen in the following reaction? Balance 12.3 Limiting Reagent and Percent Yield

Example: If the actual yield is 10.5g of NH 3 what is the percent yield? 12.3 Limiting Reagent and Percent Yield

Download ppt "Starter S-125 1.2 moles NaC 2 H 3 O 2 are used in a reaction. How many grams is that?"

Similar presentations