How many grams of nh3 can be produced from the reaction of 28 g of n2 and 25 g of h2

Chemistry

1 answer:

vfiekz [6]3 years ago

5 0

Answer:

34 g of NH₃ will be produced from the reaction of 28 g of N₂ with 25 g of H₂.

Explanation:

1) Balanced chemical equation

N₂ (g) + 3H₂ (g) → 2NH₃(g)

2) Stoichiometric (theoretical ) mole ratios

1 mol N₂ (g) : 3mol H₂ (g) : 2 mol NH₃(g)

3) Number of moles of each reactant

number of moles = mass in grams / atomic mass

number of moles of N₂ = 28 g / 28 g/mol = 1 mol

number of moles of H₂: 25 g / 2 g/mol = 12.5 mol

4) Limiting reactant

Since the stoichiometry states that 1 mol of N₂ reacts with 3 moles of H₂, the given mass of N₂ will react completely with the given amount of H₂, and the calculations must be done with the 28 g (1 mol) of N₂ as the limiting reactant.

5) Yield

Set the proportion with the mole ratios:

1 mol H₂ / 2 mol NH₃ = 1 mol H₂ / x ⇒ x = 2 mol NH₃

6) Convert to grams

mass in grams = number of moles × molar mass = 2 mol × 17 g/mol = 34 g.

Answer: the reaction of 28 g of N₂ with 25 g of H₂ will produce 34 g of NH₃

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