1.0 g of copper was heated in a crucible, while stirring, and .2g of unreacted copper was recovered usoing hydrochloric acid as a solvent, how much copper would precipitate when zinc is later added to the copper oxide/acid solution?
To determine how much copper would precipitate when zinc is added to the copper oxide/acid solution, we first need to understand the chemical reactions involved and the stoichiometry of these reactions.
1. The initial reaction is the heating of copper in air, leading to the formation of copper oxide (CuO):
2Cu + O2 → 2CuO
2. The next reaction involves the dissolution of copper oxide in hydrochloric acid (HCl):
CuO + 2HCl → CuCl2 + H2O
Now, to calculate the amount of copper that would precipitate when zinc is added, we need to consider the following reaction:
3. Zinc (Zn) reacts with copper chloride (CuCl2), resulting in the formation of copper (Cu) and zinc chloride (ZnCl2):
Zn + CuCl2 → Cu + ZnCl2
Given that 0.2g of unreacted copper was recovered using hydrochloric acid as a solvent, we can assume that the remaining copper reacted with oxygen to form copper oxide (CuO). So, the initial mass of copper would be 1.0g - 0.2g = 0.8g.
Now let's calculate the stoichiometry between copper (Cu) and copper oxide (CuO):
2Cu + O2 → 2CuO
From this equation, we can see that 2 moles of copper (2Cu) react to form 2 moles of copper oxide (2CuO). Therefore, the molar ratio between copper and copper oxide is 1:1.
Since the molar mass of copper (Cu) is approximately 63.5 g/mol, the initial amount of copper (in moles) can be calculated as follows:
Amount of copper (in moles) = Initial mass of copper (in grams) / Molar mass of copper
Amount of copper (in moles) = 0.8g / 63.5 g/mol ≈ 0.0126 mol
Now, since the reaction between zinc and copper chloride (CuCl2) has a 1:1 ratio, the amount of copper (in moles) that would precipitate when zinc is added is also 0.0126 mol.
Finally, we can calculate the mass of copper that would precipitate by multiplying the amount of copper (in moles) by the molar mass of copper:
Mass of copper that would precipitate = Amount of copper (in moles) × Molar mass of copper
Mass of copper that would precipitate = 0.0126 mol × 63.5 g/mol ≈ 0.8001 g
Therefore, approximately 0.8001 grams of copper would precipitate when zinc is added to the copper oxide/acid solution.