worksheet_05_103

CHEMISTRY 103 - WORKSHEET #5 — Module 4 Stoichiometry (Part II) Do the topics appropriate for your course Prepared by Dr. Tony Jacob https.//clc.chem.wisc.edu (Resource page) CHEMICAL REACTIONS: reactants (the starting chemicals) yield products (the ending chemicals) STOICHIOMETRY atoms or atoms or molecules A molecules B 1 mol A=6.022x 1023 A 1 mol B=6.022x 1023 B chemical formula or chemical reaction moles A |- > moles B molar mass A molar mass B grams A grams B Example 1 How many grams N»(g) are needed to prepare 250.0g N»O3(g)? 2N»s(g) +305(g) — 2N,O3(g) Answer 1: 92.1g NH3 This is a “grams A — grams B” calculation; it requires 3 steps/conversions (the steps are shown in the parentheses) * | Imol N,O 2mol N 28.0g N 2. gaflmolA (mOIB)( gb )= B 3. 250.0¢g N,O 2-3 2 2 |=92.1g NH ], 2Ny +30, —> 2N,0; 7 8 ( ¢A \mola \ImolB) ~ %" 7 ®7273( 76.0g N,05 )| 2mol N,05 )| TmoI N, 5703 LIMITING REAGENTS: one reactant runs out first — this is the limiting reagent; (How to identify this type of problem: a limiting reagent problem has 2 reactant quantities given in the problem) Example 2: Which reagent 1s the limiting reagent when 4mol N5(g) is combined with Smol O5(g)? 2N»(g) +305(g) — 2N,03(g) Answer 2: Oy(g) This question can be done with 2 calculations. Calculate the number of moles N»O5(g) that can be produced from each reactant; whichever reactant produces the smaller amount of N»O3(g) is the limiting reagent. | Step 1 Step 1: mol Ny — mol N,O3 Step 2: mol Oy — mol N,O3 2mol N,O 2mol N,O 2N + 30, == 2N:05 4mol Ny| 0—223 | =4mol N,O3 Smol Oy ———2-3 |=3 3mol N,O;4 | 2mol N, 3mol O, Step 2 O»(g) 1s the LR since it produced the smaller amount of N»O3(g); 3.3mol N5O3 from O, versus 4mol N5O3 from Ny

Example 3: a. How many grams Al,O3(s) can be made from 25.0g Al(s) and 20.0g O5(g) using: 4Al(s) + 305(g) — 2A1,053(s)? b. Which reagent is the limiting reagent? c. Which reagent is the excess reagent? d. How much of the excess reagent remains after the reaction is complete? Answer 3: This problem can be identified as a limiting reagent problem because there are 2 reactant quantities were given in the problem and can be solved by doing several “grams A — grams B” calculations and comparing the results. Step 1 Imol Al [ 2mol Al,O 102.0g Al,O Step 1: gAl —> gAl,05: 25.0g Al| — 273 52073 | 2 47.2g ALO;;: | Y 27.0g Al 4mol Al Imol Al,O4 4A1 + 30, — 2 Al,0;4 Step 2: g0, — gAl,03: 20.0g O, =42.5g Al,04 Step 3 Step 2 32.0¢0, 3mol O, Imol Al,O4 Calculate the amount of product that can be produced from each reactant (see Steps 1 and 2). Several questions can now be answered. a. 42.5g Al,O3(s) (the smaller quantity of Al,O3(s) is how much can theoretically be produced) b. O(g) is the LR (the LR reagent is the chemical that produces the smaller amount of Al,03(s)) c. Al(s) is the excess (EX) reagent (the EX reagent is the chemical that produces the larger amount of Al,O5(s)) d. Amount Excess Reagent Left Over = Starting Amount Excess Reagent — Amount Excess Reagent Used, Amount Excess Reagent Used is calculated with a third “grams A — grams B” calculation (Step 3 below); LR — EX: Imol O 4mol Al |{ 27.0g Al . Step 3: gO5 —> gAl (used): 20.0g 02( 2 ]( oo ]( g J =22.5gAl; 22.5g Al is the Amount Excess Reagent Used 32.0g O, ){ 3mol O, )\ Imol Al Amount Excess Reagent Left Over = 25.0g Al (starting amount) — 22.5g Al (amount used) = 2.5g Al left over actual yield theoretical yield PERCENT YIELD = X 100% (the “actual yield” is usually given in the problem) EMPIRICAL FORMULA from mass CO, and H,O (combustion) A. Compound contains C and H only 1. Convert gCO, = mol CO, - mol C 2. Convert gH,O — mol H,O — mol H 3. Write formula and divide by smallest moles 4. If needed, fractions: 1/2 (0.5) > x2; 1/3 or 2/3 (0.33,0.66) > x 3; 1/4 or 3/4 (0.25,0.75) > x 4 B. Compound contains C, H, and X Convert gCO, — mol CO, - mol C; mol C - gC (need both mol C and gC) Convert gH,O — mol H,O — mol H; mol H — gH (need both mol H and gH) Calculate gX from: total g sample = gC + gH + gX (gX = total g sample - gC - gH) Convert gX — mol X Write formula and divide by smallest moles If needed, fractions: 1/2 (0.5) > x2; 1/3 or 2/3 (0.33,0.66) > x 3; 1/4 or 3/4 (0.25,0.75) > x 4 AN

3. Consider the reaction below. When 4 moles of sulfur react with 6 moles of oxygen gas which (S statement is CORRECT? If all statements are incorrect, select answer “e”. Sg(g) + 80,(g) — 8SO;(g) a. O,(g) 1s the limiting reagent and Sg(g) is the excess reagent. b. Sg(g) is the limiting reagent and O,(g) is the excess reagent. c. Both O,(g) and Sg(g) are limiting reagents. d. Without knowing how much product is created it is not possible to determine the limiting reagent. ¢. None of the above are correct. 4. When 27.5g H, reacted with excess Cl, the percent yield was 75.0%. What was the actual yield of HCI1? (Hint: start by writing out the percent yield equation) Hy(g) + Cly(g) — 2HCI(g) a. 20.6g HCI b. 746g HCI c. 1930g HCI d. 1985g HCI €. 2647g HCI 5. When 5.000g of 1,5-hexadiene, a compound containing only carbon and hydrogen, is combusted 16.07g CO, and 5.483g H,O are recovered. What is the empirical formula for 1,5-hexadiene? a. CH b. CH, C. C2H d. C2H3 c. C3H5

5 6. When 5.000g of a compound containing carbon, hydrogen, and nitrogen is combusted, 13.910g CO, and 2.848g H,O are recovered. What 1s the empirical formula of the compound? a. CHN b. C4HyNy; c. CsHsN d. C1oHsN, €. C4gHgNpy