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Firstly, a balanced equation has to be written for the production of ammonia (NH₃) from hydrogen gas (H₂) and nitrogen gas (N₂):
N₂ + 3H₂ → 2NH₃
Now, the mole ratio of N₂ : NH₃ is 1 : 2 based on the coefficients of the balanced equation.
If the moles of N₂ = 2.5 moles
then the moles of NH₃ produced = 2.5 mol × 2
= 5 mol
Thus, the moles of ammonia produced when 2.5 mol of nitrogen gas is combined with excess hydrogen gas is 5 mol.
Answer:
4552 mL
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V₁) = 55 mL
Molarity of stock solution (M₁) = 12 M
Molarity of diluted solution (M₂) = 0.145 M
Volume of diluted solution (V₂) =?
The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
12 × 55 = 0.145 × V₂
660 = 0.145 × V₂
Divide both side by 0.145
V₂ = 660 / 0.145
V₂ ≈ 4552 mL
Thus, the volume of the diluted solution is 4552 mL
