Question 1 Prelaboratory Assignment: Analysis of an Aluminum-Zinc Alloyh2On the following page, construct a graph of N₁2 vs % Al. To do this, refer to equation 5 ande discussion preceding it. Calculate N₁2 for % Al =0, 25, 50, 75, and 100. Place these pointsthe graph and Draw the best straight line between them.h2A student obtained the following information from this experiment. Complete the chart:Mass of CapsuleCapsule + AlloyMass of AlloyMass of EmptyBeakereaker + Waterlass of Waterume of Water0.1168 g0.3522 g•2354 gTemperatureTemperature KBarometricPressureCVapor Pressure ofWater (Table 2)Pressure of H₂141.2 g388.3 g247.1813256 mLVolume H₂= Volume displaced water =Moles H₂ evolved (equation 6)Litersmoles273+221 20294 K751 mm Hg1.8.68/mm Hg•9636 atm (Eq. 5)NH2(g) = ((% Al/100) x 0.0556) + ((100-%A1/100) x 0.0153)

Volume of hydrogen = Volume of displaced water = 13256 ml = 13.256 lts (1000 ml = 1 lt) Pressure of hydrogen = 0.9636 atm Ideal gas constant, R = 0.0821 lt atm/mol K Temperature = 294 K According to ideal gas equation, PV = nRT 0.9636 atm x 13.256 lts = n x 0.0821 lt atm/mol K x 294 K 12.77 moles/24.14 = n n = 0.529 moles Moles of Hydrogen evolved = 0.529 moles 1) Al = 0 % NH2 = ((%Al/100) x 0.0556) + (100-%Al/100) x 0.0153) NH2 = ((0/100) x 0.0556) + (100-0/100) x 0.0153) =0.0153 moles 2) Al = 25% NH2 = ((%Al/100) x 0.0556) + (100-%Al/100) x 0.0153) NH2 = ((25/100) x 0.0556) + (100-25/100) x 0.0153) NH2 = ((25/100) x 0.0556) + (100-25/100) x 0.0153) = (0.0556/4) + (0.75 x 0.0153) = 0.0139 + 0.011475 = 0.025375 moles 3)Al = 50% NH2 = ((%Al/100) x 0.0556) + (100-%Al/100) x 0.0153) NH2 = ((50/100) x 0.0556) + (100-50/100) x 0.0153) NH2 = ((50/100) x 0.0556) + (100-50/100) x 0.0153) = (0.0556/2) + (0.5 x 0.0153) = 0.0278 + 0.00765 = 0.03545 moles 5) Al = 75% NH2 = ((%Al/100) x 0.0556) + (100-%Al/100) x 0.0153) NH2 = ((75/100) x 0.0556) + (100-75/100) x 0.0153) NH2 = ((75/100) x 0.0556) + (100-75/100) x 0.0153) = (0.0556x0.75) + (0.25 x 0.0153) = 0.0417 + 0.003825 = 0.045525 moles 6)Al = 100% NH2 = ((%Al/100) x 0.0556) + (100-%Al/100) x 0.0153) NH2 = ((100/100) x 0.0556) + (100-100/100) x 0.0153) NH2 = ((100/100) x 0.0556) + (100-100/100) x 0.0153) = (0.0556) + (0 x 0.0153) = 0.0556 +0 = 0.0556 moles