Question

1. So let's say you have a beaker that contains 12.0g of sodium hydroxide and to this you add 350mL of 0.500 M ammonium sulfate creates sodium sulfate, water and ammonia gas according to the following chemical equation: (NH4)2SO4(s) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(1) + 2NH3(g) AH = +61.12 kJ

A) Draw the beaker at the end of the reaction - what would e in the beaker and/or around it. What would the products look like on a molecular/atom/ion level?

B) At STP, how many L of your gas was produced? (ignore water vapor)

C) Why would water vapor pressure be involved in this reaction?

D) Assuming the aqueous volume given in the problem remains the same, what is the molarity of the two ions in the reaction at the end of the reaction?

E) How many moles of excess reactant are left over at the end of the reaction?

extra prob:

How many L of your gas be produced if the reaction was done at 25°C and 0.992 atm.

Using the amount of your reactants, how many kJ of energy are produced or absorbed?

Answer 1

Given Reaction is (NH4)2 Soy(s) +2 Naoh (aq) - Noy soulaq) + 2H2011) + 2NH3 (9) luven 129 of NaOH = 129 given 12g of NaOH = 0 az. Molaity of WHU 2 504 = 0.500M Moles in 350 mL = 0.5 x 350 = 0.175 moles OF (NH4)2SO4 ODD 40 g/mol = 0.3 moles (NH4)2 Souls) + 2 Nadh (aq) - Na,sou + 2 H2O + 2NH3 00175 moles 0.3 moles It is limiting reagent product will For Na2SO4 be decided by it . 2 moles of NaOH givest I moles of Na₂SO4 0.5 moles of NaOH give - 1X0o3 moles (i) (ii) - 0.15 mous OY 05X 40 = For tho & moles of the gives - 2 moles of H₂O 0.5 moles of H₂O geves 0.3 moles of moles → or 0.3x18 = 5.4 q (ii) For NH₃ 2 mous of WallHmoles 0.3 moles of NH₃ 2 moles of NH3 0.5 moles of NH3 (A so If reaction occur then (NH4)2S0yls) + 2 Naohlag) - Naz Soylaq) + 2H2011) +2 NH3 (9) Att= 0 0 0175 molls 0.3 moles o o o - At lend 0.025 moles o moles 0015 moles 0.3 moles 0.5 moles.

And product will look like: 2 Na tangt součag]+ 2NHhy dont H2011) + NH3(g) SO42- = 0.025 (fuwm (NH4), 504)+0,15 (fuom Na, sou) = 0.175 m Nat = 2X0115 = 0·30 moles NH" = b 025X2 = 0.05mo LLA (B) At STP P= 1 bar T= 273.15K a Using PV = nRT , For I mole Vm = Imou x 0.0821 atm XLX273,15 K 1 bar molxk = 22.7L for 0.3mous of NH3 = 22.7 x Ooh = 6081 Litres (C) water vapour pressure As the water is present in the reaction thinket would enout some pressure known as vopour Пица That can be given according to Rault's haw | Psotz Xsolvent Psolvent) Psoin= vapour lesure of solution Psolvent=P of pure solvent x solv= mole fraction Or According to Clausius-Capeymon lequation where Ti, Tz are Temperatures Y, P. aul Pluu AH is heat of vapourisation

1) Assuming volume wemains constant 350 mb Calculating molwuty of a lons at the end of on i.e of Nazt and sou2- Na2+ = 0.30 moles SO42- = $0.15 from Na₂sou only From Al part Molaunty = No Of molly x 1000 volume of solution (m2) Na2+ = 0.3 x 100 = 10.8511 M 350 Sou? = 0.15 X 1000 = 10.4285 M) 350