Stoichiometry Problems

3.66 Silicon tetrachloride (SiCl4) can be prepared by heating Si in chlorine gas:

Si(s) + 2 Cl2 (g)  SiCl4(l)

In one reaction, 0.507 mole of SiCl4 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 H2 (g) + N2 (g)  2 NH3(g) In a particular reaction, 6.0 moles of NH3 were produced. How many moles of H2 and how many moles of N2 were reacted to produce this amount of NH3?

3.68 Consider the combustion of butane (C4H10):

2 C4H10 (g) + 13 O2 (g)  8 CO2 (g) + 10 H2O(l) In a particular reaction, 5.0 moles of C4H10 are reacted with an excess of O2. Calculate the number of moles of CO2 formed.

3.71 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 0.140 g of KCN is treated with an excess of HCl, calculate the amount of HCN formed, in grams.

3.72 Fermentation is a complex chemical process of wine making in which glucose is converted into ethanol and carbon dioxide:

C6H12O6  2 C2H5OH + 2 CO2 glucose ethanol Starting with 500.4 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 = 0.789 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 + O2 + 2 H2O  4 KAu(CN) 2 + 4 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)  2Cl2(g)   SiCl4(l) Strategy: Looking at the balanced equation, how do we compare the amounts of Cl2 and SiCl4? 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 Cl2 and SiCl4 is known: 2 moles Cl2  1 mole SiCl4. From this relationship, we have two conversion factors. 2 mol Cl2 1 mol SiCl4

and

1 mol SiCl4 2 mol Cl2

Which conversion factor is needed to convert from moles of SiCl4 to moles of Cl2? The conversion factor on the left is the correct one. Moles of SiCl4 will cancel, leaving units of "mol Cl2" for the answer. We calculate moles of Cl2 reacted as follows: ? mol Cl 2 reacted  0.507 mol SiCl4 

2 mol Cl2  1.01 mol Cl 2 1 mol SiCl4

Check: Does the answer seem reasonable? Should the moles of Cl2 reacted be double the moles of SiCl4 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 H2 and N2 that reacted to produce 6.0 moles of NH3. 3 H2(g)  N2(g)  2 NH3(g)

3.68

? mol H 2  6.0 mol NH3 

3 mol H 2  9.0 mol H 2 2 mol NH3

? mol N 2  6.0 mol NH3 

1 mol N2  3.0 mol N2 2 mol NH3

Starting with the 5.0 moles of C4H10, we can use the mole ratio from the balanced equation to calculate the moles of CO2 formed. 2 C4H10(g)  13 O2(g)  8 CO2(g)  10 H2O(l) ? mol CO2  5.0 mol C4 H10 

3.71

8 mol CO2  20 mol CO2  2.0  101 mol CO2 2 mol C4 H10

The balanced equation shows a mole ratio of 1 mole HCN : 1 mole KCN. 0.140 g KCN 

1 mol KCN 1 mol HCN 27.03 g HCN    0.0581 g HCN 65.12 g KCN 1 mol KCN 1 mol HCN

3.72

C6H12O6   2 C2H5OH  2 CO2 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)  180.16 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 C2 H5OH  500.4 g C6 H12 O6 

1 mol C6 H12O6 2 mol C2 H5OH 46.07 g C2 H5OH   180.16 g C6 H12 O6 1 mol C6 H12 O6 1 mol C2 H5OH

 255.9 g C2H5OH

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 =

3.73

255.9 g = 324 mL = 0.324 L 0.789 g/mL

The mass of water lost is just the difference between the initial and final masses. Mass H2O lost  15.01 g  9.60 g  5.41 g moles of H2 O  5.41 g H 2O 

3.74

1 mol H 2O  0.300 mol H 2O 18.02 g H 2O

The balanced equation shows that eight moles of KCN are needed to combine with four moles of Au.

? mol KCN  29.0 g Au 

1 mol Au 8 mol KCN  = 0.294 mol KCN 197.0 g Au 4 mol Au