How many grams of ammonia NH3 are necessary to react with 1.2X10^24 molecules of oxygen?
To determine the number of grams of ammonia required to react with a certain number of molecules of oxygen, we need to use the concept of moles and the balanced chemical equation. Here's how you can solve the problem step by step:
Step 1: Write and balance the chemical equation for the reaction between ammonia (NH3) and oxygen (O2):
4NH3 + 5O2 → 4NO + 6H2O
Step 2: Determine the molar ratio between ammonia and oxygen. From the balanced equation, we see that 4 moles of ammonia react with 5 moles of oxygen.
Step 3: Convert the given number of oxygen molecules to moles. Avogadro's number tells us that 1 mole of a substance contains 6.022 × 10^23 molecules. Therefore, we can calculate:
1.2 × 10^24 molecules of oxygen = (1.2 × 10^24 molecules) / (6.022 × 10^23 molecules/mol) = 1.99 moles of oxygen
Step 4: Use the molar ratio from step 2 to find the number of moles of ammonia needed. We know that 4 moles of ammonia react with 5 moles of oxygen. Therefore,
4 moles NH3 / 5 moles O2 = x moles NH3 / 1.99 moles O2
Solving for x (the number of moles of NH3):
x = (4/5) × 1.99 = 1.592 moles of NH3
Step 5: Convert the moles into grams using the molar mass of ammonia. The molar mass of ammonia (NH3) is calculated by adding the atomic masses of nitrogen (N) and hydrogen (H). The atomic mass of nitrogen is 14.01 g/mol, and the atomic mass of hydrogen is 1.01 g/mol.
Molar mass of NH3 = (3 × atomic mass of H) + atomic mass of N = (3 × 1.01 g/mol) + 14.01 g/mol = 17.03 g/mol
Therefore, the number of grams of ammonia needed is:
1.592 moles NH3 × 17.03 g/mol = 27.13 grams of ammonia
So, you would need approximately 27.13 grams of ammonia (NH3) to react with 1.2 × 10^24 molecules of oxygen (O2).