How many grams of [tex]NH_3[/tex] can be produced from the reaction of 28 g of [tex]N_2[/tex] and 25 g of [tex]H_2[/tex]?

To calculate how many grams of NH3 can be produced from the reaction of 28g of N2 and 25g of H2, we need to use stoichiometry and the balanced chemical equation N2(g) + 3H2(g) → 2NH3(g). First, we convert the masses of N2 and H2 to moles using their molar masses (N2: 28.02 g/mol, H2: 2.02 g/mol).

For N2: 28 g N2 × (1 mol N2 / 28.02 g N2) = 1 mol N2
For H2: 25 g H2 × (1 mol H2 / 2.02 g H2) = 12.38 mol H2

According to the chemical equation, 3 moles of H2 are needed to produce 2 moles of NH3. Therefore, hydrogen is the excess reactant in this situation. We use the moles of nitrogen to find the amount of NH3 that can be produced since it is the limiting reactant.

1 mol N2 will produce 2 mol NH3 (as per the balanced equation).

Now, convert the moles of NH3 to grams using its molar mass (NH3: 17.03 g/mol).

2 mol NH3 × 17.03 g/mol = 34.06 g NH3

Therefore, 28 grams of N2 and 25 grams of H2 can produce 34.06 grams of NH3, provided that N2 is the limiting reactant.

Answered by PhilDavis | 2024-06-18

The reaction of 28 g of N₂ with 25 g of H₂ can produce approximately 34.06 grams of NH₃. Nitrogen is the limiting reactant in this scenario. The reaction can be understood through stoichiometric calculations based on the balanced chemical equation.
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Answered by PhilDavis | 2024-10-01