(a) State two factors which can affect the rate of a chemical reaction. (b) 0.72g of magnesium was added to different volumes of 2 mol. per dm\(^{3}\) hydro...

(a) Two factors that affect the rate of a chemical reaction: temperature and concentration of the reactants. (Surface area, a catalyst and, for gases, pressure are also acceptable.)

(b) The data to be plotted are:

Vol. of 2 mol dm-3 HCl used (cm3)	5	15	25	35	45
Vol. of H2 evolved (cm3)	120	360	550	600	600

Plot volume of H₂ (y-axis) against volume of acid (x-axis). The graph rises steeply and in direct proportion at first (about 24 cm³ of H₂ per 1 cm³ of acid), then bends over and becomes a horizontal line (a plateau) at 600 cm³ once all the magnesium has reacted.

(c) From the graph:

• (i) At 50 cm³ of acid the curve is already on the plateau, so the volume of hydrogen is 600 cm³ (adding more acid gives no more gas because the magnesium is used up).
• (ii) To produce 480 cm³ of hydrogen, read on the rising (proportional) part of the graph. Since \(120\ \text{cm}^3\) H₂ corresponds to \(5\ \text{cm}^3\) acid, the ratio is \(24\ \text{cm}^3\) H₂ per cm³ acid: \[V_{acid} = \frac{480}{24} = \mathbf{20\ cm^3}\]
• (iii) The acid needed to dissolve the magnesium exactly is the volume at which the graph just reaches the plateau (600 cm³), which is read as 30 cm³.

(d) At 30 cm³ of acid, the acid is exactly enough to react with all 0.72 g of magnesium; this is the point where the maximum volume of hydrogen (600 cm³) is first produced. Below this volume the acid is the limiting reactant (less hydrogen). Above it, the magnesium is completely used up and is now the limiting reactant, so no extra hydrogen is formed and the graph stays flat.

(e) First find the moles of magnesium used:

\[n_{Mg} = \frac{0.72}{24} = 0.03\ \text{mol}\]

(i) From (c)(iii), 0.03 mol Mg needs 30 cm³ of the acid. For 1 mole:

\[V = \frac{30}{0.03} = 1000\ \text{cm}^3 = \mathbf{1\ dm^3}\]

(ii) From the plateau, 0.03 mol Mg liberates 600 cm³ of hydrogen. For 1 mole:

\[V_{H_2} = \frac{600}{0.03} = 20000\ \text{cm}^3 = \mathbf{20\ dm^3}\]

(iii) Deduce the mole ratios. The acid is 2 mol dm^-3, so 1 dm³ contains 2 mol HCl. Therefore:

• 1 mol Mg reacts with 2 mol HCl (from (e)(i)).
• 1 mol Mg produces about 1 mol H₂ (20 dm³ is close to the molar gas volume, from (e)(ii)).

This gives the ratio Mg : HCl : H₂ = 1 : 2 : 1, so the equation is:

\[\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)\]