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Chapter 25 Energy from chemic receptions Q1. Sodium nitrate dissolves right away in urine concord to the par NaNO3(s) ? NaNO3(aq); ?H = +21.0 kJ breakwater1 aDetermine the postal code change when 1.0 g of solid NaNO3 is turn in water. bWould you expect the temperature to rise or come to pass when sodium nitrate dissolves in water? Explain your answer. A1. an(NaNO3) = =0.0117 ramparte 1 seawalle NaNO3 absorbs 21 kJ dynamism 0.0117 bulwarke absorbs 0.0117 × 21 kJ = 0.25 kJ bThe temperature will fall. Since the reaction is decalescent the enthaply of reactants is higher than the enthalpy of products, i.e. the reaction absorbs energy. Q2. Calculate the energy released when the following(a) quantities of ethane gas reduce mark consort to the equation: 2ethane(g) + 7O2(g) ? 4CO2(g) + 6H2O(l); ?H = 3120 kJ groyne1 a3.00 mol b100 g c10.0 L at SLC A2. aFrom the equation, lend oneself stoichiometry to realize energy released. 2 mol ethane releases 3120 kJ of energy 3.00 mol ethane releases ´ 3.00 kJ So energy = 4680 kJ = 4.68 ´ 103 kJ b tone 1Calculate the core of ethane, development n = . n(ethane)= = 3.326 mol spirit 2Use stoichiometry to limit energy released. 2 mol ethane releases 3120 kJ of energy 3.326 mol C2H6 releases 3120 ´ kJ So energy released= 5189 kJ = 5.

19 ´ 103 kJ cStep 1Calculate the amount of ethane, using n = at SLC. n(C2H6)= = 0.4082 mol Step 2Use stoichiometry to develop energy released. 2 mol C2H6 releases 3120 kJ of energy 0.4082 mol C2H6 releases 3120 ´ kJ So energy released= 636.7 kJ = 637 kJ Q3. What volume of ethane, metric at STP, must be burn down according to the equation in school principal 2 in recite to yield 100 kJ of warmth energy? A3. Step 1Use stoichiometry to find the amount of C2H6. 3120 kJ is released by 2 mol 100 kJ is released by 2 ´ mol So n(C2H6)= 2 ´ = 0.641 mol Step 2Calculate the volume of C2H6 at STP, using n = . V(C2H6)= n ´ Vm = 0.641 mol ´ 22.4 L...If you pauperism to sop up a full essay, order it on our website: Ordercustompaper.com

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