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QuestionMix 0.216 moles of N₂ with 0.596 moles of H₂. Find the limiting reactant and max moles of NH₃ produced.

Studdy Solution

STEP 1

Assumptions1. The reaction is given by the equation _{}(g) +3H_{}(g) \rightarrowNH_{3}(g) . The initial amount of nitrogen gas (₂) is0.216 moles3. The initial amount of hydrogen gas (H₂) is0.596 moles4. The reaction goes to completion5. The reaction follows the stoichiometry given in the equation

STEP 2

First, we need to find out which reactant is the limiting reactant. The limiting reactant is the reactant that is completely consumed first in a chemical reaction. We can do this by comparing the mole ratio of the reactants with the stoichiometry of the reaction.
The stoichiometry of the reaction tells us that1 mole of nitrogen gas reacts with moles of hydrogen gas to produce2 moles of ammonia.

STEP 3

Now, let's calculate the mole ratio of nitrogen gas to hydrogen gas in the given mixture.
Mole ratio=Moles of N2Moles of H2Mole\, ratio = \frac{Moles\, of\, N_{2}}{Moles\, of\, H_{2}}

STEP 4

Plug in the given values for the moles of nitrogen gas and hydrogen gas to calculate the mole ratio.
Mole ratio=0.2160.596Mole\, ratio = \frac{0.216}{0.596}

STEP 5

Calculate the mole ratio.
Mole ratio=0.2160.596=0.362Mole\, ratio = \frac{0.216}{0.596} =0.362

STEP 6

Compare the calculated mole ratio with the stoichiometric ratio. If the calculated mole ratio is less than the stoichiometric ratio, then nitrogen gas is the limiting reactant. If the calculated mole ratio is greater than the stoichiometric ratio, then hydrogen gas is the limiting reactant.
The stoichiometric ratio of nitrogen gas to hydrogen gas is13, or0.333.
Since the calculated mole ratio (0.362) is greater than the stoichiometric ratio (0.333), hydrogen gas is the limiting reactant.

STEP 7

Now that we know the limiting reactant, we can calculate the maximum amount of ammonia that can be produced. According to the stoichiometry of the reaction,3 moles of hydrogen gas produce2 moles of ammonia.
So, the maximum amount of ammonia that can be produced is given by the formulaMaximum moles of NH3=23×Moles of limiting reactantMaximum\, moles\, of\, NH_{3} = \frac{2}{3} \times Moles\, of\, limiting\, reactant

STEP 8

Plug in the given value for the moles of the limiting reactant (hydrogen gas) to calculate the maximum moles of ammonia.
Maximum moles of NH3=23×0.596Maximum\, moles\, of\, NH_{3} = \frac{2}{3} \times0.596

STEP 9

Calculate the maximum moles of ammonia.
Maximum moles of NH3=23×.596=.397 molesMaximum\, moles\, of\, NH_{3} = \frac{2}{3} \times.596 =.397\, molesThe limiting reactant is hydrogen gas (H₂) and the maximum amount of ammonia (NH₃) that can be produced is.397 moles.

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