Mg + 2HCl --> MgCl2 + H2 For a mass of 0.013 g Mg and 5.0 mL of 1.00 M HCl, in a
ID: 748890 • Letter: M
Question
Mg + 2HCl --> MgCl2 + H2 For a mass of 0.013 g Mg and 5.0 mL of 1.00 M HCl, in a flask with V = 140.612 mL at 23.5?C. Calculate the initial and final pressure (atm).Explanation / Answer
lets firts calculate the moles intially mole of Mg = wt/M.wt = 0.013/24 = 5.41 * 10^-4 molarity of HCl = 1 => mole = molarity * volume = 1 * 0.140612 = 0.14 0612 total mole initially = 0.140612 + 0.000541 = 0.14115 now initial pressure P = nRT/V = 0.14115 * 0.0821 * 296.5/0.140612 = 24.5 atm now Mg + 2HCl --> MgCl2 + H2 since one mole of Mg reacts with 2 mole of HCl and yields one mole of MgCl2 and one mole of H2 0.000541 mole of Mg reacts with 0.00108 mole of HCl now total mole in reaction mixutre is 0 + (0.140612 - 0.00108) + (0.000541 + 0.00541) = 0.140612 hence pressure = 0.140612 * 0.0821 * 296.5/ 0.140612 = 24.3 atm