1. based on the equation below, how many liters of water can be produced from 8.5 L (assume that hydrogen is in excess)
2H2(g)+O2(g) --->2H2O(g)
When using volume only with gases, one can ignore going through the conversion to mols and just take the ratio as the volumes in the equation. You read the equation this way. 2 volumes of hydrogen react with 1 volume of oxygen. Therefore, 2.O L of oxygen at STP would react with ?? L of hydrogen at STP
To determine the number of liters of water that can be produced from 8.5 L of hydrogen gas, you need to use the coefficient ratio between the reactants and the product in the balanced chemical equation.
From the equation, we can see that 2 moles of hydrogen gas (H2) react with 1 mole of oxygen gas (O2) to produce 2 moles of water (H2O).
1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 L. So, 2 moles of water will occupy 44.8 L (2 x 22.4 L).
Now, you can set up a proportion to determine the volume of water produced from 8.5 L of hydrogen gas:
(8.5 L H2) / (2 mol H2) = (V L H2O) / (2 mol H2O)
Simplifying the proportion:
(8.5 L / 2 mol) = (V L / 2 mol)
V L = (8.5 L * 2 mol) / 2
V L = 8.5 L
Therefore, based on the given equation, 8.5 L of water can be produced from 8.5 L of hydrogen gas when hydrogen is in excess.