Question

DATA Trial 1 Trial 2 Mass of Mg ribbon (g) TO SU 0.07839 (Proom) Barometric pressure (mmHg) 756 matte 756rrotta | 22.4°C Temperature of water inside beaker (C) 23.0°C 22.8°C Temperature of the room (°C) (T2) Pwater (mm Hg) Vapor pressure of water found in table at the end of the experiment, based on the temperature of water inside the beaker. 5 mm Hg Volume from graduated cylinder (mL) (V2) 60.2mL 82.4mL CALCULATIONS . Write the balanced chemical equation for the reaction between Mg and HCI. • Calculate the number of moles of Mg that were used for each trial.

Answer 1

Date: Anda). The teaction taking place between My & Hel is - Mg (g) + Helag) Moch + H2 (9) - 6 Now balancing the above equation we get Mg(s) +2H ce (aq) ? Mg Clat Halg)/ In the first trial we know that most of Mg ribbon used is 0.05419 Similarly, we know the amount of made used in the decand trial ie 0.07839. Now, we know that the atomie mass of Mq is 2 24.304 g Lo, moles wed in first trial = 0.054 I = 0.0022 24.304 Limilarly, for second trial = 0.0783 -0.0032 24. 304 So, moles of Mo (In finst Trial) = 0.0022 moles (In Second trials 0.0032 moles. From the eq n (i), we know that the H2(g) is released in this reaction. Now, this H2(g) is reacted further I with the atomopherie oogger resulting a formation water the tecection is carried as follows Hi (g) + 1 O₂(g) + H2O (ag) Here the no. of moles of water (H2O(g) formed is equal to the no. of moled of Hily)

Page : Date: Lo, we can assume to find the moles of H2 (9) place of water as the volatile liquid (water) i jormed from vapours ie Applying Ideal Gas low, PV = nRT. When, P = atom, v = litres, R=0.082 atm t - 'Kand T = Kelvin

Date: / / In first trial, Broom= 756 mmHg, Perater = 21.068m V 60.2 ml = 0.0602 l, R=0.082/afmelk inalt T= 23 °C = 296K ie 1 atom= 760mm Hg. Hence Preg = 734 932 = 0.96 atm 7GO Now, abblying PV = nRT , we get 0.96 80.0002=nxo.0821 X 2 9 6. = 0.058 = nx 22.085 In= 0.026 moles of Hz Limilarly, In gecond trial Prega Proom- Pwater: 756-20.815=735.185 mm Hg V = 82.4 ml = 0.08246 T= 22.800 = 295.8K je PV = nRT * 0.967 0.0824=nX 0.0821 x 295.8 n = 0.0797 = 0.0328 24.28 ic In=0.0328 mala of Hil d) The Barometric pressure ie Proom is equal in both the trials ie first & decond. - 750mm Hg Now, we know that I atm = 700 minta Hence, 756 mm Hg = 756 10. 995 atm