DAT - Dental Admission General Chemistry: Stoichiometry Questions and Answers 1 — Questions and Answers
Question 1: In the complete combustion of butane (C4H10), what is the sum of the stoichiometric coefficients of the reactants and products in the balanced chemical equation?
- 15
- 33 (Correct answer)
- 18
- 21
Correct answer: 33
The balanced chemical equation for the complete combustion of butane is 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O. To find the sum of the coefficients, you add them together: 2 + 13 + 8 + 10 = 33. It's crucial to balance the equation first, starting with carbons, then hydrogens, and finally oxygens.
Question 2: A compound is found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. If the molar mass of the compound is approximately 180 g/mol, what is its molecular formula?
- C2H4O2
- CH2O
- C6H12O6 (Correct answer)
- C3H6O3
Correct answer: C6H12O6
First, assume a 100g sample, which gives 40.0g C, 6.7g H, and 53.3g O. Convert these masses to moles: C (40.0g / 12.01 g/mol ≈ 3.33 mol), H (6.7g / 1.01 g/mol ≈ 6.63 mol), O (53.3g / 16.00 g/mol ≈ 3.33 mol). Divide by the smallest number of moles (3.33) to get the empirical formula CH2O. The molar mass of the empirical formula (CH2O) is about 30 g/mol. To find the molecular formula, divide the given molar mass by the empirical formula mass (180 g/mol / 30 g/mol = 6). Multiply the subscripts in the empirical formula by this factor (6) to get C6H12O6.
Question 3: Consider the reaction: 2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g). If 10.8 g of aluminum (Al) reacts with 36.5 g of hydrochloric acid (HCl), which of the following statements is correct? (Molar masses: Al = 27.0 g/mol, HCl = 36.5 g/mol)
- Aluminum is the limiting reactant, and 0.6 moles of H2 are produced.
- Hydrochloric acid is the limiting reactant, and 0.5 moles of H2 are produced. (Correct answer)
- Aluminum is the limiting reactant, and 1.0 mole of H2 is produced.
- Both reactants are completely consumed.
Correct answer: Hydrochloric acid is the limiting reactant, and 0.5 moles of H2 are produced.
First, convert the mass of each reactant to moles: Moles of Al = 10.8 g / 27.0 g/mol = 0.4 mol. Moles of HCl = 36.5 g / 36.5 g/mol = 1.0 mol. According to the stoichiometry, 2 moles of Al react with 6 moles of HCl, a 1:3 ratio. For 0.4 mol of Al, 1.2 mol of HCl would be needed (0.4 * 3). Since only 1.0 mol of HCl is available, HCl is the limiting reactant. Now, calculate the moles of H2 produced from the limiting reactant: 1.0 mol HCl * (3 mol H2 / 6 mol HCl) = 0.5 mol H2.
Question 4: In a laboratory experiment, the reaction of 32.5 g of zinc with excess hydrochloric acid produced 0.90 g of hydrogen gas. What is the percent yield of hydrogen? The balanced equation is: Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g). (Molar masses: Zn = 65.4 g/mol, H2 = 2.0 g/mol)
- 90.0% (Correct answer)
- 45.0%
- 99.4%
- 89.5%
Correct answer: 90.0%
First, calculate the theoretical yield of H2. Convert the mass of Zn to moles: 32.5 g Zn / 65.4 g/mol ≈ 0.497 mol Zn. According to the 1:1 mole ratio between Zn and H2, 0.497 moles of H2 should be produced. Convert moles of H2 to grams: 0.497 mol H2 * 2.0 g/mol = 0.994 g H2 (theoretical yield). The actual yield is given as 0.90 g. The percent yield is (Actual Yield / Theoretical Yield) * 100 = (0.90 g / 0.994 g) * 100 ≈ 90.5%, which is closest to 90.0%.
Question 5: Which of the following correctly defines the empirical formula of a compound?
- The formula that shows the actual number of atoms of each element in a molecule.
- The formula representing the simplest whole-number ratio of atoms of each element in the compound. (Correct answer)
- A formula that can only be determined from the molar mass of the compound.
- The formula that illustrates the three-dimensional arrangement of atoms in a molecule.
Correct answer: The formula representing the simplest whole-number ratio of atoms of each element in the compound.
The empirical formula represents the simplest whole-number ratio of elements within a compound. For example, the molecular formula for glucose is C6H12O6, but its empirical formula is CH2O, as the ratio 6:12:6 can be simplified to 1:2:1.
Question 6: The synthesis of ammonia is represented by the equation: N2(g) + 3H2(g) → 2NH3(g). How many grams of hydrogen gas (H2) are required to produce 68 grams of ammonia (NH3)? (Molar masses: N = 14.0 g/mol, H = 1.0 g/mol)
- 6 g
- 34 g
- 12 g (Correct answer)
- 4 g
Correct answer: 12 g
First, calculate the molar mass of NH3: 14.0 + 3(1.0) = 17.0 g/mol. Then, convert the mass of NH3 to moles: 68 g NH3 / 17.0 g/mol = 4.0 moles of NH3. According to the balanced equation, 3 moles of H2 are required for every 2 moles of NH3 produced. So, moles of H2 needed = 4.0 moles NH3 * (3 moles H2 / 2 moles NH3) = 6.0 moles of H2. Finally, convert moles of H2 to grams. The molar mass of H2 is 2(1.0) = 2.0 g/mol. Mass of H2 = 6.0 moles H2 * 2.0 g/mol = 12 g.
DAT - Dental Admission General Chemistry: Stoichiometry Questions and Answers 1
In the complete combustion of butane (C4H10), what is the sum of the stoichiometric coefficients of the reactants and products in the balanced chemical equation?