Balance the equation
NH3 + O2 −→ N2 + H2O
Given 4.55 mol of the reactant NH3, determine the corresponding amount of O2.
2NH3 + O2 → N2 + 3H2O
Note that you need 2 moles of NH3 for each mole of O2
So, you'd need 2.275 mol of O2
To balance the given chemical equation, NH3 + O2 → N2 + H2O, we need to make sure that the same number of atoms of each element is present on both sides of the equation.
Let's start by balancing the nitrogen (N) atoms. There are 1 nitrogen atom on each side of the equation, so we don't need to make any changes.
Next, let's balance the hydrogen (H) atoms. There are 3 hydrogen atoms on the left side (3 in NH3) and 2 hydrogen atoms on the right side (2 in H2O). To balance this, we can multiply H2O by 3 to give us 3 H2O on the right side.
The equation now becomes:
NH3 + O2 → N2 + 3 H2O
Now let's balance the oxygen (O) atoms. There are 3 oxygen atoms on the right side (3 in 3 H2O), but none on the left side. To balance this, we need to add a coefficient of 3 in front of O2 on the left side.
The balanced equation is:
NH3 + 3 O2 → N2 + 3 H2O
Now, to determine the corresponding amount of O2 consumed when 4.55 mol of NH3 is reacted, we can use the balanced equation to set up a mole-to-mole ratio.
According to the equation, 1 mol of NH3 reacts with 3 mol of O2. Therefore, we can set up the following proportion:
(4.55 mol NH3) / (1 mol NH3) = (x mol O2) / (3 mol O2)
Cross-multiplying and solving for x, we have:
x = (4.55 mol NH3) * (3 mol O2) / (1 mol NH3)
x = 13.65 mol O2
Therefore, the corresponding amount of O2 consumed when 4.55 mol of NH3 is reacted is 13.65 mol.