Chemical Reactions and Stoichiometry Flashcards

In the reaction 2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O, if 0.50 mol of KMnO₄ reacts completely, how many grams of Cl₂ gas are produced? (Molar mass of Cl₂ = 71 g/mol)

Answer: 17.75 g

From the balanced equation, 2 mol KMnO₄ produces 5 mol Cl₂. So 0.50 mol KMnO₄ produces (0.50 × 5/2) = 1.25 mol Cl₂. Mass = 1.25 × 71 = 88.75 g — wait, that's choice D. Let me re-examine: 0.50 mol KMnO₄ × (5 mol Cl₂ / 2 mol KMnO₄) = 1.25 mol Cl₂. 1.25 × 71 = 88.75 g. The correct answer is 88.75 g.

A reaction has a theoretical yield of 45.0 g but only 36.9 g of product is recovered. What is the percent yield, and what term describes the leftover starting material if one reactant was completely consumed?

Answer: 82% yield; the remaining reactant is the excess reagent

Percent yield = (actual/theoretical) × 100 = (36.9/45.0) × 100 = 82%. The reactant that is NOT fully consumed — the one still remaining after the reaction — is called the excess reagent. The one completely consumed is the limiting reagent.

Iron(III) oxide reacts with carbon monoxide: Fe₂O₃ + 3CO → 2Fe + 3CO₂. If 80.0 g of Fe₂O₃ (MM = 160 g/mol) reacts with 60.0 g of CO (MM = 28 g/mol), which is the limiting reagent and how many grams of Fe (MM = 56 g/mol) are produced?

Answer: CO is limiting; 40.0 g Fe

Moles Fe₂O₃ = 80/160 = 0.500 mol; needs 3 × 0.500 = 1.500 mol CO. Moles CO available = 60/28 = 2.143 mol. Since only 1.500 mol CO is needed but 2.143 mol is available, Fe₂O₃ might not be limiting. Check from CO side: 2.143 mol CO × (1 mol Fe₂O₃/3 mol CO) = 0.714 mol Fe₂O₃ needed, but only 0.500 mol available — so Fe₂O₃ is limiting. Fe produced = 0.500 mol Fe₂O₃ × (2 mol Fe/1 mol Fe₂O₃) = 1.000 mol Fe = 56 g. The correct answer is B: Fe₂O₃ is limiting; 56.0 g Fe.

Consider the combustion of butane: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O. At STP, what volume of CO₂ (in liters) is produced when 29.0 g of C₄H₁₀ (MM = 58 g/mol) completely combusts? (1 mol of gas = 22.4 L at STP)

Answer: 89.6 L

Moles C₄H₁₀ = 29.0/58 = 0.500 mol. From the equation, 2 mol C₄H₁₀ produces 8 mol CO₂, so the ratio is 4 mol CO₂ per mol C₄H₁₀. CO₂ produced = 0.500 × 4 = 2.00 mol. Volume = 2.00 × 22.4 = 44.8 L — wait, that's choice A. Let me recheck: 0.500 mol C₄H₁₀ × (8 mol CO₂ / 2 mol C₄H₁₀) = 2.00 mol CO₂. 2.00 × 22.4 = 44.8 L. The correct answer is A: 44.8 L.

A student dissolves a mixture of NaCl and an inert filler. After reacting the mixture with excess AgNO₃, 2.870 g of AgCl precipitate forms (MM of AgCl = 143.5 g/mol; MM of NaCl = 58.5 g/mol). What mass of NaCl was in the original mixture?

Answer: 1.170 g

The reaction is NaCl + AgNO₃ → AgCl + NaNO₃ (1:1 mole ratio). Moles AgCl = 2.870 / 143.5 = 0.02000 mol. Since moles NaCl = moles AgCl (1:1 ratio), moles NaCl = 0.02000 mol. Mass NaCl = 0.02000 × 58.5 = 1.170 g.

In a double displacement reaction, 50.0 mL of 0.200 M Pb(NO₃)₂ is mixed with 30.0 mL of 0.500 M KI. The reaction is Pb(NO₃)₂ + 2KI → PbI₂↓ + 2KNO₃. How many moles of PbI₂ precipitate form?

Answer: 0.00750 mol

Moles Pb(NO₃)₂ = 0.0500 L × 0.200 mol/L = 0.0100 mol. Moles KI = 0.0300 L × 0.500 mol/L = 0.0150 mol. For complete reaction, 0.0100 mol Pb(NO₃)₂ needs 2 × 0.0100 = 0.0200 mol KI, but only 0.0150 mol KI is available — KI is limiting. Moles PbI₂ = 0.0150 mol KI × (1 mol PbI₂ / 2 mol KI) = 0.00750 mol.