the reaction of N2+3H2-->2NH3 is used to produce ammonia commercially. if 1.40 g of N2 are used in the reaction, how many grams of H2 will be needed?

To find out how many grams of H2 are needed, we first need to determine the stoichiometry of the reaction, which is the balanced equation:

N2 + 3H2 --> 2NH3

From the equation, we can see that for every 1 mole of N2, we need 3 moles of H2.

To calculate the grams of H2 needed, we can follow these steps:

1. Convert the mass of N2 from grams to moles.
- To do this, we need to know the molar mass of N2, which is approximately 28 g/mol.
- Divide the given mass of N2 (1.40 g) by the molar mass of N2:
1.40 g N2 / 28 g/mol N2 = 0.05 mol N2

2. Use the stoichiometric ratio to determine the moles of H2 needed.
- According to the balanced equation, for every 1 mole of N2, 3 moles of H2 are needed.
- Multiply the moles of N2 by the stoichiometric ratio:
0.05 mol N2 × 3 mol H2 / 1 mol N2 = 0.15 mol H2

3. Convert moles of H2 to grams of H2.
- To do this, we need to know the molar mass of H2, which is approximately 2 g/mol.
- Multiply the moles of H2 by the molar mass of H2:
0.15 mol H2 × 2 g/mol H2 = 0.30 g H2

Therefore, 0.30 grams of H2 will be needed to react with 1.40 grams of N2.

To determine the grams of H2 needed, we first need to calculate the mole ratio between N2 and H2 based on the balanced chemical equation:

N2 + 3H2 -> 2NH3

From the equation, we can see that for every 1 mol of N2, we need 3 mol of H2.

Now, let's calculate the moles of N2:

Molar mass of N2 = 14.007 g/mol × 2 = 28.014 g/mol

Moles of N2 = Mass / Molar mass = 1.40 g / 28.014 g/mol = 0.05 mol

Using the mole ratio between N2 and H2, we can calculate the moles of H2:

Moles of H2 = 0.05 mol N2 × (3 mol H2 / 1 mol N2) = 0.15 mol

To convert the moles of H2 to grams, we can use the molar mass of H2:

Molar mass of H2 = 1.008 g/mol × 2 = 2.016 g/mol

Grams of H2 = Moles × Molar mass = 0.15 mol × 2.016 g/mol ≈ 0.30 g

Therefore, approximately 0.30 grams of H2 will be needed for the reaction.

More of the same. All of these stoichiometry problems are done the same way.