If you had 78.2 grams of K, how much oxygen is needed to convert it all to K2O?
78.2g K = 2 moles
each 2 moles of K reacts with 1 mole of O.
1 mole O = 16g
78
To determine the amount of oxygen needed to convert 78.2 grams of K (potassium) to K2O (potassium oxide), you first need to understand the chemical equation for this reaction.
The balanced equation for the reaction between potassium and oxygen to form potassium oxide is:
4K + O2 -> 2K2O
From the equation, we can see that 4 moles of potassium react with 1 mole of oxygen to produce 2 moles of potassium oxide. To find the amount of oxygen required, we will follow these steps:
Step 1: Calculate the molar mass of potassium (K).
The molar mass of potassium (K) is 39.10 g/mol.
Step 2: Convert the given mass of potassium (78.2 grams) to moles.
To do this, divide the given mass by the molar mass:
Moles of K = 78.2 g / 39.10 g/mol = 2 moles
Step 3: Use the mole ratio from the balanced equation to determine the moles of oxygen required.
From the balanced equation, we know that 4 moles of potassium react with 1 mole of oxygen. So, if 2 moles of potassium are reacting, we will need (2 moles K) x (1 mole O2 / 4 moles K) = 0.5 moles of oxygen.
Step 4: Convert moles of oxygen to grams by multiplying by the molar mass of oxygen (O2).
The molar mass of oxygen is 32.00 g/mol.
Mass of O2 = (0.5 moles) x (32.00 g/mol) = 16.0 grams
Therefore, to convert 78.2 grams of K to K2O, you would need approximately 16.0 grams of oxygen.