Cs+H2O=>CsOH+H2. A 0.5 kilogram block of cesium is dropped into the ocean. How much gas is released into the atmosphere if the air temp is 12 C?

Answer 1

THis is a combination problem -

First it's a straight up stoichiometry problem.

Then it's a gas-law problem.

So let's look at the stoichiometry first.

The given reaction is:

Cs (s) + H₂O (l) --> CsOH (aq) + H₂ (g)

Always, always, ALWAYS, check if the reaction is balanced. This one is not. SO let’s balance it.

2Cs (s) + 2H₂O (l) --> 2CsOH (aq) + H₂ (g)

Now, check the question of limiting reagent. In this problem, we’re reacting 0.5 kg of cesium with a whole OCEAN of water. I think we can be sure that the water is in excess; the cesium is limiting.

Therefore, we will calculate the moles of H₂ gas generated on the basis of how much Cesium we started with:

0.5 kg * (1000 g/ kg) * (1 mol Cs/ 132.9 g) * (1 mol H₂/ 2 mol Cs) = 1.88 mol

So now we know "how much" Hydrogen gas is formed - - - it's 1.88 moles!

However, you must always check what your problem wants. Does the problem want "How much" in terms of grams? Or liters? For grams, you would convert from mole to grams using the molar mass.

But I think this one is liters, because it's a gas.

SO let's use PV=nRT to figure out how many liters of gas is 1.88 mol H₂ at a temperature of 12 C.

P = atmospheric pressure, 1.00 atm

V = ??? this is the question!

n = 1.88 moles, which we figured out in the first part of the question.

R = gas constant! Choose the right value! We are going to use the value that has the units in atmospheres and liters and Kelvin: 0.08206 L*atm/K*mol

T = 12 C, which is 285 Kelvin. (DON'T forget to change temperature to Kelvin when doing gas law calculations!)

Plug the values into the equation and solve:

1.00 atm * V = 1.88 mol * .0821 L*atm/K*mol * 285 K

V = 43.989 LIters

3 sig figs would be 44.0 L

1 sig fig would be 40 L.