The balanced reaction for the production of water from hydrogen and oxygen is: 2H₂ + O₂→ 2H₂O. Also, the relevant molar masses are the following:
Molar mass of H₂O = 18 g/mol
Molar mass of O₂ = 32 g/mol
We need not be concerned with the amount of hydrogen since we can derive the amount of oxygen that reacted based on the amount of product produced. The following equation is then set-up:
Mass of O₂ reacted = <span>202 g H2O x 1 mol
H2O/ 18 g H2O x 1 mol O2/ 2 mol H2O x 32 g O2/ 1 mol O2 = 179.56 g O2
Thus, we have determined that 179.56 g O2 reacted with H2 to produce 202 g of H2O.</span>
The supposition is substantial on the grounds that, for these whose enthalpies is more prominent than n-pentane, they have a positive ΔH(polar), which implies there are more polar than the n-pentane. for these whose enthalpies is littler than n-pentane, they have a negative ΔH(polar), which implies there are less polar than the n-pentane. This supposition has an impact when I consider the predominant IMFS and different IMFS in the outline segment.
Answer: depending on the method, 138-139 (imperial) gallons, or 626 L
Explanation:
The wt of water is 4924-3547 lb = 1377 lb = 1377/2.2 kg = 626 kg = 626 L
1gallon = 4.5 L (it does where I come from and who still measures things in imperial anyway?) so I guess that is 139 gallons
or alternatively, I recall from distant childhood, “a pint of water weighs a pound and a quarter” which means 1 gallon = 10 lb, so 1377 lb = ~138 gallons
