If 37.4 grams of water decomposes at 297 Kelvin and 1.30 atmospheres, how many liters of oxygen gas can be produced? Show all of the work used to solve this problem.

2 H2O (l) 2 H2 (g) + O2 (g)

Question

If 37.4 grams of water decomposes at 297 Kelvin and 1.30 atmospheres, how many liters of oxygen gas can be produced? Show all of the work used to solve this problem. 

2 H2O (l)  2 H2 (g) + O2 (g)

anonymous · Answer

answer is 19,4 dm3

anonymous · Answer

Atomic weights: H=1, O=16, H2O=18, O2=32 

37.4gH2O x 1molH2O/18gH2O x 1molO2/2molH2O x 22.7LO2/1molO2 = 23.6 L at STP

Second step: P1V1T1 = P2V2/T2, so V2 = P1V1T2/T1P2. And then V2 = (1atm)(23.6L)(297K)/(273K)(1.30atm) = 19.7 L to three significant figures

anonymous · Answer

n(H2O)=m(H2O)/M(H2O)=2,07 mol
n(O2)= 1/2 * n(H2O) = 1,035 mol
pV=nRT --> V=nRT/p
V(O2)= (1,035 mol * 8,314 JK-1mol-1 * 297K)/131722,5 Pa
V(O2) = 0,0194 m3 --> 19,4 dm3 or 19,4 liters