Question

4. (2) How many grams of calcium sulfate will form from the reaction of 25.0 mL of 0.35 M potassium sulfate solution with excess calcium nitrate? Write the balanced equation first. (Solution Stoichiometry 101) 5. (0.5 p each) Classify the following processes as exothermic or endothermie. Condensation of 12: Sublimation of carbon dioxide: 6. (1 p) What is the specific heat of a metal rod that weighs 250 g and increases its temperature from 22.5 °C to 54.7 °C when 2400 J of heat are transferred to it? 7. (1 p) What is the heat associated with the complete reaction of 25 mols of ammonia? 4 NH3(g) + 5 O2(g) - 4 NO(g) + 6 H2O(g) AH = -906 kJ

Answer 1

4) Potassium sulfate + Calcium nitrate calcium sulfate + Potassium nitrate > Casou +2KNO3 Ca(NO3)2 excess K2SO4 + 0.35M O 25 ml 8.75 mmol 8-75X10-3 mol Give Imol of casou - so, Imol of k₂sou 8.758 ios mol of k2sou give 8.75 X 10 mal of Casoy amount of X 136,14 casou = 8.75 X 10 =1.19 g amount of casoy formed is 1.19 g | Condensation of I2 I, (e) I, (s) DHE-ve This is an exothermic process sublimation of CO₂ (g) co2 2 CO2 (s) DH = - ve This is also exothermic. and since, in both process, Molecule hence force of attraction increase released. so both condensation and are exothermic. comes closer and heat sublimation process w= 250g = 0.25 kg T = 22.5° c T2 = 54.7°C DE = 2400 I

Specific heat =- DE m DT DE m (I-1) 2400 3 0.25 kg (54.7 -22.5) 'c = 22 lege 2400 0.25 X 32. Specific hext = 298.13 - 7 4 Nha (9) + 5O2 (g) - 4 NO (g) + 6420 (g) DH = -906 kj 4 mol of 25 mol of Nna who (ammonial give give 906 ki heat a 966 x 25 u --5,662.5 kg Note : -ve sign indicate release of heat.