46g of Sodium reacted with 18g of Oxygen. How much Sodium Oxide would be produced?

To solve this problem, we need to use the concept of stoichiometry, which is a section of chemistry that studies the quantitative relationships between reactants and products in a chemical reaction.

Step 1: Write down the balanced chemical equation for the reaction. 2Na + O2 → 2Na2O

Step 2: Calculate the molar mass of each reactant. The molar mass of Sodium (Na) is approximately 23 g/mol and the molar mass of Oxygen (O2) is approximately 32 g/mol.

Step 3: Convert the mass of each reactant to moles. For Sodium: 46 g ÷ 23 g/mol = 2 mol For Oxygen: 18 g ÷ 32 g/mol = 0.5625 mol

Step 4: Determine the limiting reactant. From the balanced chemical equation, we know that the reaction requires 2 moles of Sodium for every 1 mole of Oxygen. Since we have more than enough Sodium (2 moles) for the amount of Oxygen we have (0.5625 mol), Oxygen is the limiting reactant.

Step 5: Use the stoichiometry of the reaction to determine the amount of Sodium Oxide produced. From the balanced chemical equation, we know that 1 mole of Oxygen will produce 2 moles of Sodium Oxide. Therefore, 0.5625 mol of Oxygen will produce 0.5625 * 2 = 1.125 mol of Sodium Oxide.

Step 6: Convert the amount of Sodium Oxide from moles back to grams. The molar mass of Sodium Oxide (Na2O) is approximately 62 g/mol. Therefore, 1.125 mol of Sodium Oxide is equivalent to 1.125 * 62 = 69.75 g.

So, 69.75g of Sodium Oxide would be produced.