Answer:- More methanol is produced.
Explanations:- From Le Chatelier's principle,"If a change is made at equilibrium then the reaction shifts to the direction in which the change is canceled."
For example, if the temperature is increased then reaction moves to the direction where the heat is consumed means the increase in temperature favors endothermic reaction.
The given reaction is exothermic in nature. So, the decrease in temperature favors the forward reaction and the production of methanol is increases.
M of water = 2(1) + 16 = 18 g/mol
n of water = 24g/(18g/mol) = 4/3 mol
n of H2O : n of O2 = 2 : 1
2:1 = 4/3 : x
x = (4/3)/2 = 2/3 mol of O2
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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
Answer:
The answer will be 160 we have to just multiply it ok