Question

Magnesium metal reacts with hydrochloric acid to give hydrogen gas. Mg + 2 HCl → MgCl2 + H2 Use the following reaction information to answer the subsequent questions. Show all of your work. 92 mm Length of Magnesium ribbon: Barometric Pressure: 762.9 mmHg Water Temperature: 29 °C Initial Gas Volume: 0.00 mL Final Gas Volume: 67.35 mL 0.7559 Mass = Length (in mm) x (1000 mm) 1. Magnesium is usually found in the laboratory as a ribbon strip. Find the mass of the Mg strip used in this reaction: 2. Calculate the number of moles of Mg reacted. 3. How many moles of H2 gas were produced (Hint: Use the mole ratio)? 4. Calculate the pressure of the dry hydrogen gas collected. 5. Calculate the experimental molar mass of H2 given that the density of H2 is 0.0899g/L.

Answer 1

- mass of my ribbon = length (in mm) x ( 01559) = 92mm X 1000 min 0.7554 TV00 mm = 0.069469 2 ng molen mass = 24.31 g/mol. s! 0.06946g my = 24.314/mi) 0.06946 = 2.857x10-3 mol 3. From the giren ggr - O ng + zuer - mylize the 1. So, from I moke my , we get 1 mok. He su for 2.85710 mote of an 2.857410-3 motes the sunt number of males of t = 2.85 of Ely = 2.8578103 noke 4. NOW, %7-35 ml : 07-35 X10L The 29°C. (273+29) 4= 302 RE 0-082 katin-k-mol-1 n = 2.857 x 103 mol pv= nRT 2-857x80-mof x 0-087Kemalist p= net/v: są, = - = pressure of dry hydrogen x 302X 67-35 X10-32 1.05 alm. posgtx 760 mm 798 mm Hq gey collected is 798 mmity. The

5. M Nou n = 2.857X100 mola = 2.857 X10-3 mol M we know, w = Vx d = 67.35810-34 x 01089996 = 6105 5x1039 = 285 7x10-3 mol 6. 055X20-3g 2.857x10-3 mol м M = 2.119 g/mol - 2.119 g/mol su experimental enoler man