222.6 g of Sodium combine completely with 77.4 g of Oxygen.
Wondering what the question is....
Looks like 5Na2O
In order to determine the amounts of elements that combine completely in a chemical reaction, we need to use the concept of stoichiometry. Stoichiometry allows us to determine the proportions in which different reactants combine and the amounts of products formed.
To find out how much Sodium combines completely with Oxygen, we need to write a balanced chemical equation for the reaction. The balanced equation will tell us the mole ratio between Sodium and Oxygen.
The balanced equation for the reaction between Sodium and Oxygen is:
4Na + O2 -> 2Na2O
From the balanced equation, we can see that 4 moles of Sodium react with 1 mole of Oxygen to form 2 moles of Sodium Oxide.
Now, let's calculate the number of moles of Sodium and Oxygen present in the given masses:
Mass of Sodium = 222.6 g
Molar mass of Sodium (Na) = 22.99 g/mol
Number of moles of Sodium = Mass of Sodium / Molar mass of Sodium
= 222.6 g / 22.99 g/mol
= 9.67 mol
Mass of Oxygen = 77.4 g
Molar mass of Oxygen (O2) = 32.00 g/mol
Number of moles of Oxygen = Mass of Oxygen / Molar mass of Oxygen
= 77.4 g / 32.00 g/mol
= 2.42 mol
According to the balanced equation, the mole ratio between Sodium and Oxygen is 4:1. Therefore, for every 4 moles of Sodium, we need 1 mole of Oxygen.
Since we have 9.67 moles of Sodium and 2.42 moles of Oxygen, we can determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.
In this case, Oxygen is the limiting reactant because 2.42 moles of Oxygen correspond to 2.42 / 1 (moles of Oxygen / moles of Sodium) = 2.42 * 4 = 9.68 moles of Sodium that are required to react completely.
Therefore, 2.42 moles of Oxygen will combine completely with 9.68 moles of Sodium.
To find the mass of Sodium Oxide formed, we need to use the stoichiometry of the balanced equation:
Mass of Sodium Oxide = Molar mass of Sodium Oxide * Number of moles of Sodium Oxide
= (22.99 g/mol + 16.00 g/mol) * 9.68 mol
= 377.20 g
Therefore, 222.6 g of Sodium combines completely with 77.4 g of Oxygen to form 377.20 g of Sodium Oxide.