Question

If 83.6 grams of H2 and 257 grams of N2 react, how many grams of ammonia will be produced?

Answer 1

The mass of ammonia that would be produced is 312.5 g

First, we will write a balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

3H₂(g) + N₂(g) → 2NH₃(g)

This means

3 moles of hydrogen gas reacts with 1 mole of nitrogen gas to produce 2 moles of ammonia.

First, we will determine the number of moles of each reactant present

For Hydrogen (H₂)

Mass = 83.6 g

Molar mass = 2.016 g/mol

Using the formula

$Number\ of\ moles = \frac{Mass}{Molar\ mass}$

Number of moles of H₂ present = $\frac{83.6}{2.016}$

∴ Number of moles of H₂ present = 41.468254 moles

For Nitrogen (N₂)

Mass = 257 grams

Molar mass = 28.0134 g/mol

∴ Number of moles of N₂ present = $\frac{257}{28.0134}$

Number of moles of N₂ present = 9.174181 moles

Since,

3 moles of hydrogen gas reacts with 1 mole of nitrogen gas to produce 2 moles of ammonia

Then,

27.522543 moles of hydrogen gas will react with the 9.174181 moles of nitrogen gas to produce 18.348362 moles of ammonia

∴ 18.348362 moles of ammonia will be produced during the reaction

Now, for the mass of ammonia that would be produced

From the formula

Mass = Number of moles × Molar mass

Molar mass of ammonia = 17.031 g/mol

Mass of ammonia that would be produced = 18.348362 × 17.031

Mass of ammonia that would be produced = 312.49095 g

Mass of ammonia that would be produced ≅ 312.5 g

Hence, the mass of ammonia that would be produced is 312.5 g

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