Moles Calculation Example: Nickel (II) Oxide Production

How do we calculate the mass of nickel (II) oxide produced when nickel reacts with oxygen gas?

We are given the masses of nickel and oxygen gas involved in the reaction. What is the balanced chemical equation for this reaction?

Answer:

In this scenario, 155 g of nickel reacts with 39 g of oxygen gas to form nickel (II) oxide without any nickel or oxygen left over. To calculate the mass of nickel (II) oxide produced, we can follow these steps:

1. Find the number of moles of nickel and oxygen gas using their respective molar masses.
2. Use the mole ratios from the balanced chemical equation to determine the moles of nickel (II) oxide produced.
3. Finally, calculate the mass of nickel (II) oxide produced using its molar mass.

When 155 g of nickel reacts with 39 g of oxygen gas, the balanced chemical equation for the reaction is:

4 Ni + 3 O₂ → 2 Ni₂O₃

From the equation, we can see that 4 moles of nickel react with 3 moles of oxygen gas to produce 2 moles of nickel (II) oxide. This gives us the mole ratios needed to calculate the amount of product formed.

To find the number of moles of nickel and oxygen gas:

• Moles of Ni = 155 g / 58.6934 g/mol = 2.64 mol
• Moles of O₂ = 39 g / 32 g/mol = 1.22 mol

Using the mole ratios, we determine the moles of nickel (II) oxide produced:

• Moles of Ni₂O₃ = 2.64 mol / 2 = 1.32 mol
• Moles of Ni₂O₃ = 1.22 mol × (2/3) = 0.81 mol

Finally, we calculate the mass of nickel (II) oxide produced:

Mass of Ni₂O₃ = 0.81 mol × 74.6928 g/mol = 60.48 g

Therefore, 60.48 g of nickel (II) oxide is produced in the reaction.