Question

Write a balanced equation for the preparation of \(\mathrm{H}_{2}\) using (a) \(\mathrm{Mg}\) and an acid, (b) carbon and steam, (c) methane and steam.

Short answer

(a) Mg + 2HCl → H2 + MgCl2 (b) C + H2O → H2 + CO (c) 3CH4 + 3H2O → 3H2 + 3CO

Step-by-step solution

(a) Magnesium and an acid

First, we need to identify the reactants for this reaction. Magnesium (Mg) will react with hydrochloric acid (HCl), a strong acid, to produce hydrogen gas (H2) and magnesium chloride (MgCl2). Step 1: Write the unbalanced equation: Mg + HCl → H2 + MgCl2 Step 2: Balance the equation: Since there are 2 chlorine atoms in MgCl2 but only one in HCl, we need to balance the chlorines by adding a coefficient of 2 in front of HCl. This will also balance the H atoms. The balanced equation is: Mg + 2HCl → H2 + MgCl2

(b) Carbon and steam

Carbon (C) can react with steam (H2O) to produce hydrogen gas (H₂) and carbon monoxide (CO). Step 1: Write the unbalanced equation: C + H2O → H2 + CO Step 2: Balance the equation: The equation is already balanced with equal numbers of each type of atom on both sides. So the final balanced equation is: C + H2O → H2 + CO

(c) Methane and steam

In this reaction, methane (CH4) will react with steam (H2O) to produce hydrogen gas (H₂) and carbon monoxide (CO). Step 1: Write the unbalanced equation: CH4 + H2O → H2 + CO Step 2: Balance the equation: There are 4 H atoms in CH4 and 2 in H2O, giving a total of 6 H atoms on the left. To balance the H atoms, we need 3 H2 molecules on the right, so we add a coefficient of 3 in front of H2: CH4 + H2O → 3H2 + CO Now there are 3 C atoms on the right side of the equation, but only one on the left. To balance the C atoms, we need to add a coefficient of 3 in front of CH4: 3CH4 + H2O → 3H2 + 3CO Now there are 12 H atoms on the left side and only 6 on the right side. To balance the H atoms, we need to add a coefficient of 3 in front of H2O: 3CH4 + 3H2O → 3H2 + 3CO Now the equation is balanced: 3CH4 + 3H2O → 3H2 + 3CO