Question

20. A sample of ammonia (NHa) occupies 3.00L at 25°C. What is the temperature of the gas if the volume is increased to 6.25L? (6) 21. Ammonium nitrate can decompose explosively when heated: (9) 2 NH4NO3 (s) → 2 N2 (9) + 4 H2(g) + 3 O2 (g) a) How many liters of nitrogen would be formed at 450°C and 1.00atm by the explosion of 455g of ammonium nitrate? 22. Hydrogen gas is formed (and 1 other product) when zinc reacts with HCI. Write a balanced equation for this reaction. (17) a) if 25mL of hydrogen gas were collected over water at 22°C and 756.4mmHg, how many mg of zinc were reacted? b) If 525mg of HCl are reacted with zinc, how many mL of hydrogen will be produced at STP?

Answer 1

3.00L 3.OOL z 621 K. 20) According to charles law, VAT when Pressure remains constant, for a given mass of gas il y = Constant; v = 12 T should be in kelvin scale Vi = 3:00L; T1 = (25+273) = 298 K. 12 = 6.252 ; T2 = ? 298K 6.252 ; T = 298k 6.25L - Tz = = 620.8k T2 = 620.8 K-273 = 347.8.c. ~ 348 c. Temperature of the gas = 621K (62) 348c. 2 NH Nois, 2N g) + 4h(g) + 30(9) 2 moles NH No gives 2moles N gas and share Moles of xH NO = mass - 4559 = 5.684 mol. Molar mass 80.043 g/mol moles of the formed = 5.684 mol. We know I mol N at STP has the volume 22.4L Decomposition temperature = 450°C =C450 +273) K = 723 k. We have to convert the volume of 56 684 mol N to the lade volume at the lab condition using the relation Povo = P.V Povo То Volume of 5.684 mol X at STP (273 k, latm) = 5.684 molx 22.44/mol = 127.32L (atm) x (127.32L) - (latm) Vi 273K 723k


(s) v = 127 .322 X 723K = 337.192 273K Volume of ♡ = 337.2L 22) a) zn reacts with Hel to form ance and the gas. zn + 2HCl a > ZnCl₂ aq + H2Cg Imol zn produce imol Herale 22.4L at STP. At the experimental condition, 25ml 1₂ gas is produced. Convert this volume into volume at STP. PIVI = Pove ; a= 25ml x t = 0.0254 To P2= 756.4 mm Hg TL = 22 + 273) K = 295 k. Po=760.0 mm Hg Vo = ? To = 273k. 156.4 mm Hg) (25 mi) - (760.0mm Hg) Vo 295k 273k Vo = 756.4¢ 25 x 273 ml - 23.03 ml. 295 X 760 moles of Zn reacted = 23.03 m = 0.0010 275 mol mass of Zn reacted = molx Molay mass = 0.067176 g 1000 mg = 670176 mg x 67.2 mg. produces Imolt b) From the balanced equation, 2 mol Hel moles of HCl = 525 mg. Tooo ing) = 0.0144 moles (Mass/ Me 36.46 g/mol Moles of th produced = x0.6144 = 0.0072 mol. volume of H = 0.007 2 molx 22.4 b = 0.16128 Lx 1000m! = 161.3 ml. 22.02m imol 22414ml) = 0.0010275 mol % 65-38 g/mol mo