1. Given the following chemical equation, answer the questions that follow.

Transcription

1 Problems to practice on Chapter 6 1. Given the following chemical equation, answer the questions that follow. A. Balance the equation. CH 3 OH (l) + O 2(g) CO 2(g) + H 2 O (g) Balancing Organic formulas such as this are a pain, but try and do the elements in the order of Carbon, Hydrogen and Oxygen. Carbon is equal on both sides so leave methanol and carbon dioxide at 1. CH 3 OH (l) + O 2(g) CO 2(g) + H 2 O (g) There are four hydrogens in the reactants and two in the products, so take water to a coefficient of 2. CH 3 OH (l) + O 2(g) CO 2(g) + 2 H 2 O (g) This means you have three oxygen atoms in the reactants and four in the products damn, isn t that annoying, makes you have to work for it. Try this trick, multiple the larger side of the imbalance by two. CH 3 OH (l) + O 2(g) 2CO 2(g) + 4H 2 O (g) This change then needs to be reflected in how it will affect Methanol. 2CH 3 OH (l) + O 2(g) 2CO 2(g) + 4H 2 O (g) Now look at the oxygen balance. We have 4 on the reactant side and eight on the products. We can address the deficit of four atoms by adding two more oxygen gas molecules. Voila, balanced equation 2CH 3 OH (l) + 3O 2(g) 2CO 2(g) + 4H 2 O (g) B. Given that Methanol (the hydrocarbon above) has a density of g/ml, what mass of oxygen gas would be needed to fully combust 1.5 l of methanol? Oh look, another density measure and a volume, so we know we will need the density formula.

2 Density = mass/volume But since we are given a volume and a density, we need to rearrange the equation for what we don t have, the mass. Mass = Density X Volume One more thing before we start, notice that the density and volume measures use different units (ml vs. l), we need to convert one of those to the other unit so that the units cancel. 1.5 l of methanol X (1000 ml/1 l) = 1500 ml Methanol Now we can calculate the mass of Methanol we have.?mass of methanol = 1500 ml X g/ml = 1215 g Methanol The question asks for the mass of oxygen needed to fully combust the methanol, so we need to figure out how many moles of Methanol we have (so we can compute the mole ratios compared to the balanced equation).?moles Methanol = 1215 g X (1 mole/ g) = 39.2 moles Methanol According to the balanced equation we need 3 moles of oxygen for every 2 moles of methanol, so we can use this ration to calculate the number of moles of oxygen we need.?moles O 2 = 39.2 moles Methanol X (3 moles Oxygen/2 moles Methanol) = 58.7 moles O 2 Now convert this to mass and remember that the answer should have two sig figs.?mass Oxygen = 58.7 moles X (32 g/mole) = 1900 g O 2 C. If 23.4 g of water vapor was produced during a reaction of 64.0 g of methanol, was methanol the limiting reagent? To determine if methanol is the limiting reagent we need to first determine the number of moles from each of the stated substances.?moles H 2 O = 23.4 g H 2 O X (1 mole H 2 O/ g H 2 O) = 1.30 moles?moles Methanol = 64 g Methanol X (1 mole methanol/ g methanol) = 2.00 moles The ratio of Water:Methanol is 4:2 or 2:1 (reducing the fraction). If Methanol was the limiting reagent than the ratio from the formula would equal the experimental ratio Water:Methanol = 1.30/2.00 1:2, therefore Methanol is not the limiting reagent.

3 2. During the precipitation of Silver Silicate from a mixture of potassium silicate and silver nitrate you get a percent yield of 45.5% for the silver silicate solid. If I wanted 20 g of silver silicate, how much of each reactant would I need to use to guarantee my yield? What is the skeleton chemical equation for the reaction we have described? K 4 SiO 4(aq) + Ag(NO 3 ) 2(aq) AgSiO 4(s) + KNO 3(aq) Balance the chemical equation. K 4 SiO 4(aq) + 2 Ag(NO 3 ) 2(aq) Ag 2 SiO 4(s) + 4 KNO 3(aq) You always need the balanced chemical equations when doing stoichiometry problems. Now let us look at the problem itself. We are given a percent yield, so from that and the yield we want, we can calculate the theoretical yield for the reaction needed to guarantee the yield we want. Percent Yield = 100% X (actual yield/theoretical yield) Theoretical Yield = 100% X (actual yield/percent yield) *algebraic rearrangement* Theoretical Yield = 100% X (20 g/45.5%) = g Ag 2 SiO 4 Now that we have the theoretical yield needed in grams, we have to convert it to moles (remember moles are a unit we can use with the ratios in the chemical equation).?moles Ag 2 SiO 4 = g Ag 2 SiO 4 X (1 mole/ g) = mole Ag 2 SiO 4 From the stoichiometry of the balanced equation, we can determine the amount of each reactant needed. Ag 2 SiO 4 : K 4 SiO 4 = 1:1 therefore we need mole Ag 2 SiO 4 : Ag(NO 3 ) 2 = 1:2 therefore we need mole Convert the molar amounts to mass values.? g K 4 SiO 4 = moles X ( g/mole) = 36 g K 4 SiO 4? g Ag(NO 3 ) 2 = moles X ( g/mole) = 66 g Ag(NO 3 ) 2

4 3. Consider the combustion of magnesium metal and answer the following questions. A. Write the balanced equation for the combustion of magnesium metal. All metal combustion reactions follow the same formula: metal + oxygen metal oxide Mg (s) + O 2(g) MgO (s) We know the product must be MgO since Mg is a 2+ ion and O is a 2- ion, thereby equaling to zero, the requirement of all compounds of this type. Now balance the equation. 2 Mg (s) + O 2(g) 2 MgO (s) B. If 1.24 kg of magnesium metal is combusted, what mass of oxygen is needed and what mass of product is produced? Again, we need to convert the mass to units usable for stoichiometry, which would be moles.? moles Mg = 1.24 kg X (1000g / 1 kg) X (1 mole/24.31 g) = 51.0 mol Mg Use the balanced equation to determine the number of moles of oxygen needed, then convert that to the mass of oxygen needed.?g O 2 = 51.0 mol Mg X (1 mol O 2 /2 mol Mg) X (32 g O 2 /1 mol O 2 ) = 816 g O 2 C. Consider the magnesium is being burned in a crucible and all of the product collects in the crucible, if 45 g of magnesium is combusted (and magnesium is the limiting reagent), what would the mass difference of the crucible be after the reaction? Alright, a little trickier here, what am I asking for in the question? I am asking for the mass difference of the crucible after the reaction, but what does that mean? When the metal combust in the crucible and forms magnesium oxide the product will remain in the crucible, so I am asking what is the difference between the mass of the magnesium oxide produced and the magnesium used? There are a few ways to approach this problem, but let us stick to the systematic approach (it will help you in the end to always be systematic, even if some of the steps seem trivial). The first thing I need to do is use the data I have, which is the mass of magnesium used. With this I can find the number of moles.? moles Mg = 45 g Mg X (1 mole Mg/24.31 g Mg) = 1.85 mol Mg The stoichiometry of the equation tells us that Mg and MgO are in a 1:1 ratio, so we therefore have the same number of moles of MgO when the reaction completes. So we need to determine the mass of this amount of MgO.

5 ? g MgO = 1.85 mol Mg X (1 mol MgO/1 mol Mg) X (40.31 g MgO/1 mol MgO) = 75 g MgO To determine the mass difference we simply subtract the initial mass of Mg from the final mass of MgO.?mass difference = 75 g MgO 45 g Mg = +30 g difference 4. You are given a 2.4 l bottle of 3.4 M sulfuric acid, how much NaOH do you need to add to completely neutralize the solution? Again we have a stoichiometry problem and therefore we need the balanced equation. If you read the question you will not the words acid and neutralize. These words indicate that we are performing an acid/base neutralization reaction. Let us write the skeleton chemical equation of the reaction. Now we need to balance the equation. H 2 SO 4(aq) + NaOH (aq) H 2 O (l) + Na 2 SO 4(aq) H 2 SO 4(aq) + 2NaOH (aq) 2H 2 O (l) + Na 2 SO 4(aq) To neutralize the acid means that we need to add enough NaOH to react with all the sulfuric acid present, so we have to determine the number of moles of sulfuric acid present. We start with the Molarity Equation. Molarity (M) = moles of solute (m)/volume of solution (V) Since we have the molarity and the volume, we need to solve for m. m = MV? mol H 2 SO 4 = 3.4M H 2 SO 4 X 2.4 l = 8.16 mol H 2 SO 4 Using the stoichiometry of the equation and the molar mass of NaOH we can determine the mass of NaOH needed to neutralize the raction.? g NaOH = 8.16 mol H 2 SO 4 X (2 mol NaOH/1 mol H 2 SO 4 ) X (40 g NaOH/1 mol NaOH) = 650 g NaOH