Sample Problem 7.1 Number of Atoms in Balanced Chemical Equations Indicate the number of each type of atom in the following balanced chemical equation: Fe2S3(s) + 6HCl(aq) → 2FeCl3(aq) + 3H2S(g) Solution The total number of atoms in each formula is obtained by multiplying the coefficient by each subscript in a chemical formula General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.1 Number of Atoms in Balanced Chemical Equations Continued Study Check 7.1 When ethane, C2H6, burns in oxygen, the products are carbon dioxide and water The balanced chemical equation is written as Δ 2C2H6(g) + 7O 2(g) → 4CO2(g) + 6H2O(g) Calculate the number of each type of atom in the reactants and in the products Answer In both the reactants and products, there are C atoms, 12 H atoms, and 14 O atoms General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.2 Balancing a Chemical Equation The chemical reaction of methane, CH 4, and oxygen gas, O 2, produces carbon dioxide (CO2) and water (H2O) Write a balanced chemical equation for this reaction Solution Step Write an equation using the correct formulas for the reactants and products Δ CH4(g) + O2(g) → CO2(g) + H2O(g) General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.2 Balancing a Chemical Equation Continued Step Count the atoms of each element in the reactants and products When we compare the atoms on the reactant side and the atoms on the product side, we see that there are more H atoms in the reactants and more O atoms in the products Step Use coefficients to balance each element We will start by balancing the H atoms in CH because it has the most atoms By placing a coefficient of in front of the formula for H2O, a total of H atoms in the products is obtained Only use coefficients to balance an equation Do not change any of the subscripts: This would alter the chemical formula of a reactant or product General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.2 Balancing a Chemical Equation Continued We can balance the O atoms on the reactant side by placing a coefficient of in front of the formula Step O There are now O atoms in both the reactants and products Check the final equation to confirm it is balanced In the final equation, the numbers of atoms of C, H, and O are the same in both the reactants and the products The equation is balanced General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.2 Balancing a Chemical Equation Continued In a balanced chemical equation, the coefficients must be the lowest possible whole numbers Suppose you had obtained the following for the balanced equation: Although there are equal numbers of atoms on both sides of the equation, this is not written correctly To obtain coefficients that are the lowest whole numbers, we divide all the coefficients by Study Check 7.2 Balance the following chemical equation: Al(s) + Cl2(g) → AlCl3(s) Answer 2Al(s) + 3Cl2(g) → 2AlCl3(s) General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.3 Balancing Chemical Equations with Polyatomic Ions Balance the following chemical equation: Na3PO4(aq) + MgCl2(aq)2 → Mg3(PO4)2(s) + NaCl(aq) Solution Step Write an equation using the correct formulas for the reactants and products Na3PO4(aq) + MgCl2(aq) → Mg3(PO4)2(s) + NaCl(aq) Step Unbalanced Count the atoms of each element in the reactants and products When we compare the number of ions in the reactants and products, we find that the equation is not balanced In this equation, we can balance the phosphate ion as a group of atoms because it appears on both sides of the equation General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.3 Balancing Chemical Equations with Polyatomic Ions Continued Na3PO4(aq) + MgCl2(aq) → Mg3(PO4)2(s) + NaCl(aq) Step Use coefficients to balance each element We begin with the formula that has the highest subscript values, which in this equation is Mg 3(PO4)2 The subscript in Mg3(PO4)2 is used as a coefficient for MgCl2 to balance magnesium The subscript in Mg 3(PO4)2 is used as a coefficient for Na3PO4 to balance the phosphate ion 2Na3PO4(aq) + 3MgCl2(aq) → Mg3(PO4)2(s) + NaCl(aq) General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.3 Balancing Chemical Equations with Polyatomic Ions Continued In the reactants and products, we see that the sodium and chloride ions are not yet balanced A coefficient of is placed in front of the NaCl to balance the equation 2Na3PO4(aq) + 3MgCl2(aq) → Mg3(PO4)2(s) + 6NaCl(aq) Step Check the final equation to confirm it is balanced A check of the total number of ions confirms the equation is balanced A coefficient of is understood and not usually written 2Na3PO4(aq) + 3MgCl2(aq) → Mg3(PO4)2(s) + 6NaCl(aq) General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake Balanced © 2016 Pearson Education, Inc Sample Problem 7.3 Balancing Chemical Equations with Polyatomic Ions Continued Study Check 7.3 Balance the following chemical equation: Pb(NO3)2(aq) + AlBr3(aq) → PbBr2(s) + Al(NO3)3(aq) Answer 3Pb(NO3)2(aq) + 2AlBr3(aq) → 3PbBr2(s) + 2Al(NO3)3(aq) General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.11 Mass of Product When acetylene, C2H2, burns in oxygen, high temperatures are produced that are used for welding metals Δ 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g) How many grams of CO are produced when 54.6 g of C2H2 is burned? Solution Step State the given and needed quantities (grams) Step Write a plan to convert the given to the needed quantity (grams) General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.11 Mass of Product Continued Step Use coefficients to write mole–mole factors; write molar mass factors if needed Step Set up the problem to give the needed quantity (grams) General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.11 Mass of Product Continued Study Check 7.11 Using the equation in Sample Problem 7.11, calculate the grams of CO that can be produced when 25.0 g of O reacts Answer 27.5 g of CO2 General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.12 Moles of Product from Limiting Reactant The chemical reaction of carbon monoxide and hydrogen is used to produce methanol, CH3OH CO(g) + 2H2(g) → CH3OH(g) If 3.00 moles of CO and 5.00 moles of H2 are the initial reactants, what is the limiting reactant and how many moles of methanol can be produced? Solution Step State the given and needed moles Step Write a plan to convert the moles of each reactant to moles of product General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.12 Moles of Product from Limiting Reactant Continued Step Write the mole–mole factors from the equation Step Calculate the number of moles of product from each reactant and select the smaller number of moles as the amount of product from the limiting reactant Moles of CH3OH (product) from CO: General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.12 Moles of Product from Limiting Reactant Continued Moles of CH3OH (product) from H2: The smaller amount, 2.50 moles of CH3OH, is the maximum amount of methanol that can be produced from the limiting reactant, H 2, when it is completely consumed Study Check 7.12 If the initial mixture of reactants for Sample Problem 7.12 contains 4.00 moles of CO and 4.00 moles of H 2, what is the limiting reactant and how many moles of methanol can be produced? Answer H2 is the limiting reactant; 2.00 moles of methanol can be produced General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.13 Mass of Product from a Limiting Reactant When silicon dioxide (sand) and carbon are heated, the products are silicon carbide, SiC, and carbon monoxide Silicon carbide is a ceramic material that tolerates extreme temperatures and is used as an abrasive and in the brake discs of sports cars How many grams of CO are formed from 70.0 g of SiO and 50.0 g of C? Heat SiO 2(s) + 3C(s) → SiC(s) + 2CO(g) Solution Step State the given and needed grams Step Write a plan to convert the grams of each reactant to grams of product General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.13 Mass of Product from a Limiting Reactant Continued Step Write the molar mass factors and mole–mole factors from the equation Molar mass factors: Mole–mole factors: General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.13 Mass of Product from a Limiting Reactant Continued Step Calculate the number of grams of product from each reactant and select the smaller number of grams as the amount of product from the limiting reactant Grams of CO (product) from SiO2: Grams of CO (product) from C: The smaller amount, 65.3 g of CO, is the most CO that can be produced General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.13 Mass of Product from a Limiting Reactant Continued Study Check 7.13 Hydrogen sulfide burns with oxygen to give sulfur dioxide and water How many grams of sulfur dioxide are formed from the reaction of 8.52 g of H 2S and 9.60 g of O2? Δ 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g) Answer 12.8 g of SO2 General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.14 Calculating Percent Yield On a space shuttle, LiOH is used to absorb exhaled CO from breathing air to form LiHCO3 LiOH(s) + CO2(g) → LiHCO3(s) What is the percent yield of LiHCO for the reaction if 50.0 g of LiOH gives 72.8 g of LiHCO 3? Solution Step State the given and needed quantities Step Write a plan to calculate the theoretical yield and the percent yield Calculation of theoretical yield: General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.14 Calculating Percent Yield Continued Calculation of percent yield: Step Write the molar mass factors and the mole–mole factor from the balanced equation Molar mass factors: Mole–mole factors: General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.14 Calculating Percent Yield Continued Step Calculate the percent yield by dividing the actual yield 1given2 by the theoretical yield and multiplying the result by 100% Calculation of theoretical yield: Calculation of percent yield: A percent yield of 51.3% means that 72.8 g of the theoretical amount of 142 g of LiHCO was actually produced by the reaction Study Check 7.14 For the reaction in Sample Problem 7.14, what is the percent yield of LiHCO if 8.00 g of CO2 produces 10.5 g of LiHCO 3? Answer 84.7% General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.15 Calculating Heat in a Reaction How much heat, in kilojoules, is released when nitrogen and hydrogen react to form 50.0 g of ammonia? N2(g) + 3H2(g) → 2NH3(g) ΔH = –92.2 kJ Solution Step State the given and needed quantities Step Write a plan using the heat of reaction and any molar mass needed Step Write the conversion factors including heat of reaction General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc Sample Problem 7.15 Calculating Heat in a Reaction Continued Step Set up the problem to calculate the heat Study Check 7.15 Mercury(II) oxide decomposes to mercury and oxygen 2HgO(s) → 2Hg(l) + O2(g) ΔH = +182 kJ a Is the reaction exothermic or endothermic? b How many kilojoules are needed to react 25.0 g of mercury(II) oxide? Answer a endothermic b 20.5 kJ General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc ... products Answer In both the reactants and products, there are C atoms, 12 H atoms, and 14 O atoms General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson... numbers of atoms of C, H, and O are the same in both the reactants and the products The equation is balanced General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake ©... treat skin conditions such as acne, psoriasis, and dandruff Answer 138.12 g General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C Timberlake © 2016 Pearson Education, Inc