suppose 2.5x10^4g of nitrogen gas and 5x10^5g of hydrogen are mixed and reacted to form ammonia, what is the excess reagent
To determine the excess reagent in a chemical reaction, we need to compare the amounts of the reactants and their stoichiometry (molar ratios) in the balanced chemical equation. Let's start by writing the balanced equation for the reaction between nitrogen gas (N₂) and hydrogen gas (H₂) to form ammonia (NH₃):
N₂ + 3H₂ → 2NH₃
From the balanced equation, we can see that one mole of nitrogen gas reacts with three moles of hydrogen gas to produce two moles of ammonia.
Step 1: Convert the given masses of nitrogen and hydrogen to moles.
- Nitrogen (N₂):
- Given mass: 2.5 × 10^4 grams
- Molar mass of N₂: 28 g/mol
- Moles of nitrogen: (2.5 × 10^4 g) / (28 g/mol) = 892.86 moles
- Hydrogen (H₂):
- Given mass: 5 × 10^5 grams
- Molar mass of H₂: 2 g/mol
- Moles of hydrogen: (5 × 10^5 g) / (2 g/mol) = 2.5 × 10^5 moles
Step 2: Determine the limiting reagent.
The limiting reagent is the reactant that is completely consumed in the reaction. It determines the maximum amount of product that can be formed.
To find the limiting reagent, we need to compare the moles of nitrogen and hydrogen based on the chemical equation's stoichiometry.
According to the balanced equation, 1 mole of nitrogen reacts with 3 moles of hydrogen. Therefore, if there are fewer moles of hydrogen than three times the moles of nitrogen, hydrogen will be the limiting reagent.
- Moles of Hydrogen × 3 = (2.5 × 10^5 moles) × 3 = 7.5 × 10^5 moles
Since 7.5 × 10^5 moles of hydrogen is greater than the actual amount of hydrogen present (2.5 × 10^5 moles), the moles of hydrogen present will be completely consumed. Hence, the hydrogen gas is the limiting reagent.
Step 3: Calculate the excess reagent.
To determine the excess reagent, subtract the moles of the limiting reagent consumed from the total moles of the excess reagent initially present.
- The moles of nitrogen initially present: 892.86 moles (calculated earlier)
- The moles of nitrogen consumed: 892.86 moles (since nitrogen is not in excess)
- The moles of nitrogen remaining: 892.86 moles - 892.86 moles = 0 moles
Since there are zero moles of nitrogen remaining, it means that all the nitrogen was consumed in the reaction. Hence, nitrogen gas is the excess reagent in this case.
Therefore, the excess reagent is nitrogen gas.