What is the maximum mass, in grams, of NH3 that can be produced by the reaction of 1.0 g of N2 and 1.0 g of H2 using the reaction below?
N2+H2→NH3 (not balanced)
What is the maximum mass, in grams, of that can be produced by the reaction of 1.0 g of and 1.0 g of using the reaction below?
(not balanced)
A) 0.60 g
B) 5.7 g
C) 1.2 g
D) 2.9 g
1.2
To find the maximum mass of NH3 that can be produced by the reaction, we need to follow a few steps.
Step 1: Calculate the number of moles of N2 and H2 using their respective molar masses.
First, we need to calculate the number of moles of N2:
Molar mass of N2 = 2(14.01 g/mol) = 28.02 g/mol
Number of moles of N2 = Mass of N2 / Molar mass of N2 = 1.0 g / 28.02 g/mol = 0.0357 mol
Next, we calculate the number of moles of H2:
Molar mass of H2 = 2(1.008 g/mol) = 2.016 g/mol
Number of moles of H2 = Mass of H2 / Molar mass of H2 = 1.0 g / 2.016 g/mol = 0.496 mol
Step 2: Write a balanced equation for the reaction.
The given equation is: N2 + H2 → NH3
In order to balance this equation, we need to make sure that the number of atoms on the reactant side is equal to the number of atoms on the product side.
The balanced equation is: N2 + 3H2 → 2NH3
Step 3: Determine the limiting reactant.
To find the limiting reactant, we need to compare the moles of N2 and H2 and identify which one is present in a smaller amount.
From the balanced equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Since the mole ratio between N2 and H2 is 1:3, we need 1/3 moles of N2 to react with 1 mole of H2 to produce 2/3 moles of NH3.
However, we have 0.0357 moles of N2 and 0.496 moles of H2.
Comparing 0.0357 moles of N2 and 0.496 moles of H2, we can see that N2 is the limiting reactant because it is present in a smaller amount.
Step 4: Calculate the maximum mass of NH3 produced.
From the balanced equation, we know that 1 mole of N2 produces 2 moles of NH3.
Number of moles of NH3 produced = Number of moles of N2 × (2 moles of NH3 / 1 mole of N2) = 0.0357 mol × (2/1) = 0.0714 mol
Now, we can calculate the maximum mass of NH3 produced:
Molar mass of NH3 = 2(1.008 g/mol) + 14.01 g/mol = 16.03 g/mol
Mass of NH3 produced = Number of moles of NH3 produced × Molar mass of NH3 = 0.0714 mol × 16.03 g/mol = 1.145 g
Therefore, the maximum mass of NH3 that can be produced by the reaction is 1.145 g.
Given the answer choices, the answer closest to 1.145 g is option C) 1.2 g.