2 . Question is Part b The equation below describes the combustion of propane. (

Question

2.       Question is Part b

The equation below describes the combustion of propane.  (5 points)

C3H8(g) + 5 O2(g)

Question is Part b The equation below describes the combustion of propane. C3H8(g) + 5 O2(g) à 3 CO2(g) + 4 H2O (l) In the presence of excess oxygen, how many g of CO2 are formed if 322 g of propane are consumed? (322 g) (1 mol / 44.0 g) = 7.318 mol C3H8 (7.318 mol C3H8) (3 CO2) / (1 C3H8) = 21.95 mol CO2 (21.95 mol CO2)(44.0 g) / ( 1 mol) = 966 g CO2 Using your answer from question a, what volume of CO2 would be produced at 1 atm and 22DegreeC?????

Explanation / Answer

pv=nrt

1v=nrt

v=(966/44)*0.0821*295=531.72 liter

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