1. A 0.70 g mixture of Cu and CuO was treated with hydrogen at elevated temperature. CuO was reduced to Cu and water was formed.
(a) Write a balanced equation for the reaction. (2 marks)
CuO + H2 ==> Cu + H2O
To write a balanced equation for the reaction, let's first identify the reactants and products involved:
Reactants:
- Copper (Cu)
- Copper(II) oxide (CuO)
- Hydrogen gas (H2)
Products:
- Copper (Cu)
- Water (H2O)
Now, let's write the balanced equation step by step:
1. Start by writing the unbalanced equation:
Cu + CuO + H2 → Cu + H2O
2. Next, balance the equation by adjusting the coefficients in front of the reactants and products to ensure the same number of atoms for each element on both sides of the equation.
Cu + CuO + H2 → Cu + H2O
Since there are two copper atoms on the left side and only one on the right side, we need to balance the equation further:
2Cu + CuO + H2 → 2Cu + H2O
Now, the equation is balanced, and the coefficients indicate the number of moles or molecules of each compound or element involved in the reaction.
The balanced equation for the reaction is:
2Cu + CuO + H2 → 2Cu + H2O