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)
(b) Before the experiment, calculate the minimum amount (in mole) of hydrogen required
to ensure the reduction is completed. (3 marks)
(c) After passing with in excess amount of hydrogen gas, 0.58 g of solid remained.
Calculate the percentage of CuO in the mixture.
I showed you how to do part a last night. You should have posted the entire problem then.
(a) To write a balanced equation for the reaction, we need to understand the reaction taking place. CuO is being reduced to Cu by hydrogen gas (H2), and water (H2O) is being formed as a product.
The balanced equation for the reaction can be written as follows:
CuO + H2 -> Cu + H2O
(b) To determine the minimum amount of hydrogen required to complete the reduction, we need to use stoichiometry and the molar masses of CuO and H2.
The molar mass of CuO is calculated by adding the atomic masses of copper (Cu) and oxygen (O).
Cu = 63.55 g/mol and O = 16.00 g/mol
Molar mass of CuO = 63.55 g/mol + 16.00 g/mol = 79.55 g/mol
Now, we can set up the stoichiometric equation using the balanced equation from part (a) and the molar masses of CuO and H2.
1 mole of CuO reacts with 1 mole of H2
So, the number of moles of CuO present in the mixture can be calculated by dividing the mass of CuO by its molar mass.
mass of CuO = 0.70 g
molar mass of CuO = 79.55 g/mol
Number of moles of CuO = mass of CuO / molar mass of CuO
= 0.70 g / 79.55 g/mol
Now, since the stoichiometry of the reaction is 1:1 between CuO and H2, the minimum amount of hydrogen required will be equal to the number of moles of CuO.
Therefore, the minimum amount of hydrogen required to ensure the reduction is completed is equal to the number of moles of CuO, which can be calculated as 0.70 g / 79.55 g/mol.
(c) After passing an excess amount of hydrogen gas, 0.58 g of solid remained. To calculate the percentage of CuO in the mixture, we need to calculate the mass of CuO left after the reaction.
The initial mass of the mixture = 0.70 g (given)
The mass of the remaining solid after the reaction = 0.58 g (given)
Therefore, the mass of CuO that reacted = initial mass - mass of remaining solid
= 0.70 g - 0.58 g
= 0.12 g
To calculate the percentage of CuO in the mixture, we can use the ratio of the mass of CuO that reacted to the initial mass of the mixture.
Percentage of CuO = (mass of CuO reacted / initial mass of mixture) x 100
= (0.12 g / 0.70 g) x 100
Calculate the percentage of CuO by dividing 0.12 g by 0.70 g, then multiply the result by 100 to obtain the percentage.