A reaction is carried out where 8.23 g of nitrogen gas is combined with 6.55 g o

Question

A reaction is carried out where 8.23 g of nitrogen gas is combined with 6.55 g of oxygen gas to produce NO(g). This reaction is carried out in balloon, inside a calorimeter that contains 725 g of water at 28.7°C. If the reaction between has a yield of 82.7%, what will be the final temperature of the water in the calorimeter? Assume that the heat capacity of the calorimeter, balloon, and gasses are negligible. cH2O = 4.184 J/g•K N2(g) + O2(g) 2NO(g) Hrxn =180.6 kJ

(a) 27.5°C

(b) 8.12°C

(c) -10.0°C

(d) 38.9°C

(e) 18.6°C

**solutions say the answer is (e) but not sure how they got that

Explanation / Answer

      N2(g) + O2(g) ----> 2NO(g)    Hrxn =180.6 kJ

No of moles of N2 = 8.23/28 = 0.294 mole.

No of moles of O2 = 6.55/32 = 0.205 mole.

limiting reagent is O2

No of moles of O2 reacted = 0.205*82.7/100 = 0.17 mole

No of moles of Product formed = 0.17*2 = 0.34 mole

1 mole O2   = 180.6 kJ

SO that,

0.17 mole O2 = 180.6*0.17/1 = 30.702 kj

actual amount of energy absorbed by water   = 30.702 kj

q absorbed = m*Ch2o*DT

m = mass of water = 725 grams

DT = (Tf-28.7)

CH2O = 4.184 j/g.c

30.702*10^3 = 725*4.184*(28.7-Tf)

Tf = final temperature of water = 18.57 C

answer: e) 18.6 C

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