2 C 4 H 10 (g) + 13 O 2 (g) 8 CO 2 (g) + 10 H 2 O(l)

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1 3.66 Silicon tetrachloride (SiCl 4 ) can be prepared by heating Si in chlorine gas: Si(s) + 2 Cl 2 (g) SiCl 4 (l) In one reaction, mole of SiCl 4 is produced. How many moles of molecular chlorine were used in the reaction? 3.67 Ammonia is a principal nitrogen fertilizer. It is prepared by the reaction between hydrogen and nitrogen. 3 H 2 (g) + N 2 (g) 2 NH 3 (g) In a particular reaction, 6.0 moles of NH 3 were produced. How many moles of H 2 and how many moles of N 2 were reacted to produce this amount of NH 3? 3.68 Consider the combustion of butane (C 4 H 10 ): 2 C 4 H 10 (g) + 13 O 2 (g) 8 CO 2 (g) + 10 H 2 O(l) In a particular reaction, 5.0 moles of C 4 H 10 are reacted with an excess of O 2. Calculate the number of moles of CO 2 formed When potassium cyanide (KCN) reacts with acids, a deadly poisonous gas, hydrogen cyanide (HCN), is given off. Here is the equation: KCN(aq) + HCl(aq) KCl(aq) + HCN(g) If a sample of g of KCN is treated with an excess of HCl, calculate the amount of HCN formed, in grams.

2 3.72 Fermentation is a complex chemical process of wine making in which glucose is converted into ethanol and carbon dioxide: C 6 H 12 O 6 2 C 2 H 5 OH + 2 CO 2 glucose ethanol Starting with g of glucose, what is the maximum amount of ethanol in grams and in liters that can be obtained by this process? (Density of ethanol = g/ml.) 3.74 For many years the recovery of gold that is, the separation of gold from other materials involved the use of potassium cyanide: 4 Au + 8 KCN + O H 2 O 4 KAu(CN) KOH What is the minimum amount of KCN in moles needed to extract 29.0 g (about an ounce) of gold? SOLUTIONS: 3.66 Si(s) 2Cl 2 (g) SiCl 4 (l) Strategy: Looking at the balanced equation, how do we compare the amounts of Cl 2 and SiCl 4? We can compare them based on the mole ratio from the balanced equation. Solution: Because the balanced equation is given in the problem, the mole ratio between Cl 2 and SiCl 4 is known: 2 moles Cl 2 1 mole SiCl 4. From this relationship, we have two conversion factors. 2 mol Cl2 1 mol SiCl4 and 1 mol SiCl4 2 mol Cl2

3 Which conversion factor is needed to convert from moles of SiCl 4 to moles of Cl 2? The conversion factor on the left is the correct one. Moles of SiCl 4 will cancel, leaving units of "mol Cl 2 " for the answer. We calculate moles of Cl 2 reacted as follows: 2 mol Cl2? mol Cl2reacted mol SiCl mol Cl 2 1 mol SiCl4 Check: Does the answer seem reasonable? Should the moles of Cl 2 reacted be double the moles of SiCl 4 produced? 3.67 Starting with the amount of ammonia produced (6.0 moles), we can use the mole ratio from the balanced equation to calculate the moles of H 2 and N 2 that reacted to produce 6.0 moles of NH 3. 3 H 2 (g) N 2 (g) 2 NH 3 (g) 3 mol H2? mol H2 6.0 mol NH3 9.0 mol H 2 2 mol NH3 1 mol N2? mol N2 6.0 mol NH3 3.0 mol N 2 2 mol NH Starting with the 5.0 moles of C 4 H 10, we can use the mole ratio from the balanced equation to calculate the moles of CO 2 formed. 2 C 4 H 10 (g) 13 O 2 (g) 8 CO 2 (g) 10 H 2 O(l) 8 mol CO2? mol CO2 5.0 mol C4H10 20 mol CO2 2 mol C4H mol CO The balanced equation shows a mole ratio of 1 mole HCN : 1 mole KCN. 1 mol KCN 1 mol HCN g HCN g KCN g HCN g KCN 1 mol KCN 1 mol HCN

4 3.72 C 6 H 12 O 6 2 C 2 H 5 OH 2 CO 2 glucose ethanol Strategy: We compare glucose and ethanol based on the mole ratio in the balanced equation. Before we can determine moles of ethanol produced, we need to convert to moles of glucose. What conversion factor is needed to convert from grams of glucose to moles of glucose? Once moles of ethanol are obtained, another conversion factor is needed to convert from moles of ethanol to grams of ethanol. Solution: The molar mass of glucose will allow us to convert from grams of glucose to moles of glucose. The molar mass of glucose 6(12.01 g) 12(1.008 g) 6(16.00 g) g. The balanced equation is given, so the mole ratio between glucose and ethanol is known; that is 1 mole glucose 2 moles ethanol. Finally, the molar mass of ethanol will convert moles of ethanol to grams of ethanol. This sequence of three conversions is summarized as follows: grams of glucose moles of glucose moles of ethanol grams of ethanol? g C2H5OH 1 mol C6H12O6 2 mol C2H5OH g C2H5OH g C6H12O g C 6 H 12 O 6 1 mol C 6 H 12 O 6 1 mol C 2 H 5 OH g C 2 H 5 OH Check: Does the answer seem reasonable? Should the mass of ethanol produced be approximately half the mass of glucose reacted? Twice as many moles of ethanol are produced compared to the moles of glucose reacted, but the molar mass of ethanol is about one-fourth that of glucose. The liters of ethanol can be calculated from the density and the mass of ethanol. volume mass density Volume of ethanol obtained g = = 324 ml = g/ml L 3.73 The mass of water lost is just the difference between the initial and final masses. Mass H 2 O lost g 9.60 g 5.41 g 1 mol H2O moles of H2O 5.41 g H2O mol H2O g H 2 O 3.74 The balanced equation shows that eight moles of KCN are needed to combine with four moles of Au. 1 mol Au 8 mol KCN? mol KCN 29.0 g Au = mol KCN g Au 4 mol Au

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