Finish and balance the following equation for a precipitation reaction in molecu

Question

Finish and balance the following equation for a precipitation reaction in molecular form only. evaluate phase labels: _ (NH_4)_3 PO_4 (aq) + _ Ba(NO_3)_2 (aq) rightarrow Finish and balance the following equation in BOTH molecular and ionic forms. Include phase labels:_ H_2 SO_4 (aq) + _ Ba(OH)_2 (aq) rightarrow Foe the combustion of ammonia. 2 NH_3 (g) + 7 O_2 (g) rightarrow 4 NO_2 (g) + 6 H_2 O (g) delta H = -1132 kJ How much heat would be released if 8.5 grams of ammonia (NH_3, M.W. = 17 g/mol) react? If this much heat were transferred to 1.5 kilograms of water (specific heat = 4.184 J/g middot degree C) at 22 degree C, what would its final temperature be?

Explanation / Answer

1) 2(NH4)3PO4(aq) + 3Ba(NO3)2(aq) -----> Ba3(PO4)2(s) + 6NH4NO3(aq)

2moles of Ammonium phosphate reacts with 3moles of Barium nitrate to barium phopshate precipitate and ammonium nitrate.

2) H2SO4(aq) + Ba(OH)2(aq) -------> BaSO4(s) + 2H2O(l)

3) `2NH3 + 7O2 -----> 4NO2 + 6H2O; dH = -1132 KJ

As per the equation 2 moles of NH3 on combustion relases 1132 KJ of energy that means 2*17 = 34g NH3--->1132 KJ, then 8.5g of NH3 produces,

1132*8.5/34 = -283 KJ

4) If 283 KJ of heat energy is transfered to 1.5 kg of water then the final temperature of water is,

q = mc(T

q = 283 KJ=0.283 J, m=1.5 kg = 1500g, c = 4.184 J, Initial T = 22 oC, Final temp.=?

283 = 1500*4.184*(Tfinal-22)

solving for T, Tfinal = 22.04 oC

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