Assuming all volume measurements are made at the same temperature and pressure, how many liters of water vapor can be produced when 8.23 liters of oxygen gas react with excess hydrogen gas? show all of the work used to solve this problem. balanced equation: 2h2 (g) + o2 (g) 2h2 o(g)

For the reaction 2 K + F2 --> 2 KF,
consider K atomic wt. = 39
23.5 g of K = 0.603 moles, hence following the molar ratio of the balanced equation, 0.603 moles of potassium will use 0.3015 moles of F2. (number of moles, n = 0.3015)

Now, following the ideal gas equation, PV = nRT
P = 0.98 atm
V = unknown
n = 0.3015 moles
R = 82.057 cm^3 atm K^-1mole^-1 (unit of R chosen to match the units of other parameters; see the reference below)
T = 298 K
Solving for V,
V = (nRT)/P = (0.3015 mol * 82.057 cm^3 atm K^-1 mol^-1 * 298 K)/(0.98 atm)
solve it to get 7517.6 cm^3 as the volume of F2 = 7.5176 liters of F2 gas is needed.

2. Use the formula: volume1 * concentration 1 = volume 2 * concentration 2
where, volume 1 and concentration 1 are for solution 1 and volume 2 and solution 2 for solution 2.

Solution 1 = 12.3 M NaOH solution
Solution 2 = 1.2 M NaOH solution

Solving for volume 1, volume 1 = (12.4 L * 1.2 M)/12.3 M = 0.1366 L 

Fine out you self hx

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