ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation.

N2+3H2g----- > 2NH3g.
what is the maximum mass that can be produced from a mixture of 2.00×10³gN and 8.00×10⁴H2?

I assume you mean maximum mass NH3 and I assume 8.00E4 g H2.

This is a limiting reagent problem (LR).
N2(g) + 3H2(g)----- > 2NH3(g).

1a. Convert g N2 to mols. mols = g/molar mass = ?
1b. Convert g H2 to mols the same way.
2a. Using the coefficients in the balanced equation, convert mols N2 to mols NH3. That will be mols N2 from 1a x (2 mol NH3/1 mol N2) = mols NH3 produced.
2b. Do the same and convert mols H2 to molsNH3 produced.
2c. It is likely that 2a and 2b will not be the same. If they are different, in LR problems the SMALLER number will be the correct answer for mols.
3a. Using the smaller number from 2c, convert mols NH3 to grams NH3. grams = mols x molar mass = ?

2.0468×10^73g

To find the maximum mass of ammonia that can be produced from the given reactants, we will use the concept of stoichiometry.

Step 1: Calculate the number of moles for each reactant using their given masses and molar masses.

Given:
Mass of nitrogen (N₂) = 2.00 × 10³ g
Molar mass of nitrogen (N₂) = 28.0134 g/mol

Number of moles of nitrogen (N₂):
moles of N₂ = Mass of N₂ / Molar mass of N₂
moles of N₂ = 2.00 × 10³ g / 28.0134 g/mol

Step 2: Calculate the number of moles for hydrogen (H₂) using its given mass and molar mass.

Given:
Mass of hydrogen (H₂) = 8.00 × 10⁴ g
Molar mass of hydrogen (H₂) = 2.01588 g/mol

Number of moles of hydrogen (H₂):
moles of H₂ = Mass of H₂ / Molar mass of H₂
moles of H₂ = 8.00 × 10⁴ g / 2.01588 g/mol

Step 3: Determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and restricts the amount of product formed. To find the limiting reactant, we compare the moles of each reactant to the stoichiometric ratio.

Stoichiometric ratio:
1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.

Comparing the moles of reactants:
Moles of N₂ / Stoichiometric ratio of N₂ = Moles of H₂ / Stoichiometric ratio of H₂

Step 4: Calculate the moles of ammonia (NH₃) that can be produced using the limiting reactant.

The limiting reactant is the one that produces the smaller amount of moles of product. Let's assume nitrogen (N₂) is the limiting reactant.

Moles of ammonia (NH₃) produced = Moles of N₂ × (2 moles of NH₃ / 1 mole of N₂)

Step 5: Convert the moles of ammonia (NH₃) into grams using the molar mass of ammonia.

Given:
Molar mass of ammonia (NH₃) = 17.03052 g/mol

Mass of ammonia (NH₃) produced = Moles of NH₃ × Molar mass of NH₃

Now, let's plug in the values to calculate the maximum mass of ammonia produced.

To calculate the maximum mass of ammonia (NH3) that can be produced from the given amounts of nitrogen (N2) and hydrogen (H2), you need to determine which reactant is limiting and calculate the amount of product it can produce.

Let's begin by determining the limiting reactant:

1. Start with the given masses:
- Mass of nitrogen (N2) = 2.00×10³ g
- Mass of hydrogen (H2) = 8.00×10⁴ g

2. Convert the masses to moles:
- Moles of nitrogen (N2) = mass / molar mass of N2
- Molar mass of N2 = 28.02 g/mol (2 * atomic mass of nitrogen)
- Moles of N2 = (2.00×10³ g) / (28.02 g/mol)

- Moles of hydrogen (H2) = mass / molar mass of H2
- Molar mass of H2 = 2.02 g/mol (2 * atomic mass of hydrogen)
- Moles of H2 = (8.00×10⁴ g) / (2.02 g/mol)

3. Calculate the mole ratio between N2 and H2 in the balanced equation:
- From the balanced equation: N2 + 3H2 -> 2NH3
- For every 1 mole of N2, 3 moles of H2 are needed.

4. Determine the limiting reactant:
- Compare the moles of N2 and H2 from step 2. The reactant with fewer moles is the limiting reactant.

- Moles of N2 = [(2.00×10³ g) / (28.02 g/mol)]
- Moles of H2 = [(8.00×10⁴ g) / (2.02 g/mol)]

- Compare the moles to determine the limiting reactant.

5. Calculate the moles of NH3 produced:
- Using the mole ratio from the balanced equation, 1 mole of N2 produces 2 moles of NH3.
- Moles of NH3 = (moles of N2) * (moles of NH3 / moles of N2) [Use the mole ratio from the balanced equation.]

6. Convert moles of NH3 to mass:
- Mass of NH3 = moles of NH3 * molar mass of NH3
- Molar mass of NH3 = 17.03 g/mol (1 * atomic mass of nitrogen + 3 * atomic mass of hydrogen)
- Mass of NH3 = (moles of NH3) * (17.03 g/mol)

By following these steps, you will obtain the maximum mass of ammonia (NH3) that can be produced from the given reactants.