PROBLEM 90P
Consider the chemical reaction:
2H2O(l) → 2H2(g) + O2(g)
How many moles of H2O are required to form 1.3 L of O2 at 325 K and 0.988 atm?
Solution 90P
Balanced equation:
2H2O(l) → 2H2(g) + O2(g)
2 mol H2O produce 1 mol O2
Use Gas Law Equation to calculate mol of O2
PV = nRT
0.992*1.3 = n * 0.082057*315
n = 1.2896/ 25.848
n = 0.04989 = 0.05 mol O2 produced.
From equation, this will require: 0.05*2 = 0.1mol H2O
If you require the mass of water:
Molar mass H2O = 18g/mol
0.1mol H2O = 18*0.1 = 1.8g H2O required.