The molarity of the dilution solution is 0.050 M
<h3>Further explanation</h3>Molarity is a way to express the concentration of the solution
Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution
Where
M = Molarity
n = Number of moles of solute
V = Volume of solution
Dilution formula :
M₁=6 M
V₁=12.5 ml
V₂=1.5 L=1500 ml
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 of H₂ present =
∴ 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 =
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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