What volume of hydrogen is produced at s.t.p. When 2.60 g of zinc reacts with excess HCI according to the following equation? Zn(s) + 2HCI(aq) → ZnCI 2(aq) ...

What volume of hydrogen is produced at s.t.p. When 2.60 g of zinc reacts with excess HCI according to the following equation? Zn_(s) + 2HCI_(aq) → ZnCI _2(aq) + H_2(g) [Zn = 65; 1 mole of a gas occupies 22.4dm³ at s.t.p]

Answer Details

The balanced chemical equation is: Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) First, we need to find the moles of zinc present: m(Zn) = 2.60 g M(Zn) = 65 g/mol (molar mass of Zn) n(Zn) = m(Zn) / M(Zn) n(Zn) = 2.60 g / 65 g/mol n(Zn) = 0.04 mol According to the balanced equation, 1 mole of zinc reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas. Since we have an excess of hydrochloric acid, all the zinc will react, and the amount of hydrogen produced can be calculated as follows: n(H2) = n(Zn) × 1 mol H2 / 1 mol Zn n(H2) = 0.04 mol Now we can use the ideal gas law to find the volume of hydrogen produced at STP: PV = nRT where P is the pressure (which is 1 atm at STP), V is the volume, n is the number of moles, R is the gas constant (0.08206 L·atm/mol·K), and T is the temperature (273 K at STP). Solving for V: V = nRT / P V = (0.04 mol) × (0.08206 L·atm/mol·K) × (273 K) / (1 atm) V = 0.89 L V = 0.89 dm3 (since 1 L = 1 dm3) Therefore, the volume of hydrogen produced at STP is 0.89 dm3, which is option (B).