Question

For the reaction 2H₂(g) + O₂(g) → 2H₂O(g), what volume of water vapor can be made from 100 grams of oxygen gas and an excess of hydrogen at STP? Please show work.

Answer 1

140 L

Step-by-step explanation:

Step 1: Write the balanced equation

2 H₂(g) + O₂(g) → 2 H₂O(g)

Step 2: Calculate the moles corresponding to 100 g of oxygen

The molar mass of oxygen is 32.00 g/mol.

`100g * (1mol)/(32.00g) =3.13mol`

Step 3: Calculate the moles of water vapor formed

The molar ratio of oxygen to water vapor is 1:2.

`3.13molO_2 * (2molH_2O)/(1molO_2) =6.26molH_2O`

Step 4: Calculate the volume corresponding to 6.26 moles of water vapor

1 mole of any ideal gas under STP has a volume of 22.4 L.

`6.26mol * (22.4L)/(mol) =140 L`