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

Let's solve this using molar analysis. 

First, we need to find the number of moles in 37.4 grams of water. The molecular weight of water is 36 grams per mole. 

Therefore, $$n = {m \over M}$$where n is the number of moles, m is the mass in grams, and M is the molecular weight in grams/mole. 

Let's realize that 2 moles of water creates 1 mole of oxygen. We can find the number of moles of oxygen produced by 37.4 grams of water by the following expression: $$n_{O_2} = {\rm 1 mole ~O_2 \over 2 mole~H_2O} \cdot n_{H_2O} = {\rm 1 mole ~O_2 \over 2 mole~H_2O} \cdot \left( m_{H_2O} \over M_{H_2O} \right)$$

Once we know the number of moles of oxygen that 37.4 grams of water can produce, we can use the ideal gas law to find the volume of the gas. 

$$PV = n RT$$Use$$R =0.08205746 \left[ \rm L \cdot atm \over K \cdot mole \right]$$

anonymous · Answer

Chemistry? This is Physics. Just letting you know you'd be more successful in the Chemistry forum...