Changing the surface area of the reactants
4 marks · NormalA teacher demonstrates the reaction between 3.0 g of magnesium powder and excess dilute hydrochloric acid in a conical flask. She prepares a second flask with 3.0 g of magnesium ribbon. Both are started at the same time.
Predict which flask will finish producing gas first, and explain your answer.
State the total mass of hydrogen gas produced in each flask. (Give your answer to 2 significant figures.) [M_r(H_2) = 2.0. Ar(Mg) = 24.0] Equation:
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) The flask with magnesium powder will finish first because it has a greater surface area, so the reaction is faster.
(b) 3.0 g Mg × (1 mol H₂ / 24.0 g Mg) × (2.0 g H₂ / 1 mol H₂) = 0.25 g H₂ (2 s.f.) g · must show Stoichiometry: 1 mol Mg makes 1 mol H₂
Common mistakes: Mixing up limiting reactant; Incorrect formula mass
Show mark scheme and worked solution
Marking points
S1 (a) · 1 mark · Identifies magnesium powder flask finishes first.
S2 (a) · 1 mark · Explains in terms of greater surface area leading to faster reaction.
S3 (b) · 1 mark · Sets up correct stoichiometric calculation: mass Mg to moles Mg to moles/mass H₂.
S4 (b) · 1 mark · Calculates correct final answer: 0.25 g H₂ (to 2 s.f.), for either flask. (follow-through)
Worked solution
a) The magnesium powder flask will finish producing gas first because its surface area is greater, so the rate of reaction is faster.
b) Number of moles Mg = 3.0 g / 24.0 = 0.125 mol. From the equation: 1 mol Mg → 1 mol H₂, so 0.125 mol Mg yields 0.125 mol H₂. Mass of H₂ = 0.125 mol × 2.0 g/mol = 0.25 g. Both flasks produce 0.25 g H₂.