Answer:
1.66 moles.
Explanation:
We'll begin by calculating the number of mole in 23.2 g of nitrogen gas, N2.
This is illustrated below:
Molar mass of N2 = 2x14 = 28 g/mol
Mass of N2 = 23.2 g
Mole of N2 =.?
Mole = mass /Molar mass
Mole of N2 = 23.2/28
Mole of N2 = 0.83 mole
Next, we shall determine the number of mole in 23.2 g of Hydrogen gas, H2.
This is illustrated below:
Molar mass of H2 = 2x1 = 2 g/mol
Mass of H2 = 23.2 g
Mole of H2 =?
Mole = mass /Molar mass
Mole of H2 = 23.2/2
Mole of H2 = 11.6 moles
Next, the balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
From the balanced equation above,
1 mole of N2 reacted with 3 moles of H2 to produce 2 moles of NH3.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
1 mole of N2 reacted with 3 moles of H2.
Therefore, 0.83 moles will react with = (0.83 x 3) = 2.49 moles of H2.
From the calculations made above, we can see that only 2.49 moles out of 11.6 moles of H2 is required to react completely with 0.83 mole of N2.
Therefore, N2 is the limiting reactant.
Finally, we shall determine the maximum amount of NH3 produced from the reaction.
In this case, we shall use the limiting reactant because it will give the maximum yield of NH3 since all of it is consumed in the reaction.
The limiting reactant is N2 and the maximum amount of NH3 produced can be obtained as follow:
From the balanced equation above,
1 mole of N2 reacted to produce 2 moles of NH3.
Therefore, 0.83 mole of N2 will react to produce = (0.83 x 2) = 1.66 moles of NH3.
Therefore, the maximum amount of NH3 produced from the reaction is 1.66 moles.
= 59 capsules