Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
1a. calcium chloride (CaCl2)
b. 2HCl (aq) + Ca(OH)2 (s) —> CaCl2 (aq) + 2H2O (l)
i’m not sure about the rest but i hope this helped ^^
Losing eltron is the answer
The final volume of the methane gas in the container is 6.67 L.
The given parameters;
- <em>initial volume of gas in the container, V₁ = 2.65 L</em>
- <em>initial number of moles of gas, n₁ = 0.12 mol</em>
- <em>additional concentration, n = 0.182 mol</em>
The total number of moles of gas in the container is calculated as follows;
The final volume of gas in the container is calculated as follows;
Thus, the final volume of the methane gas in the container is 6.67 L.
Learn more here:brainly.com/question/21912477
