calculate the theoretical yield of Cu when 53.2g Cu2S react with oxygen.
(balanced equation is CU2S+02>2Cu+S02)
Please help me figure this homework out. I just dnt get how to get the theoretical yeild. I want to know to be able to do the rest right. Thanks much!
To calculate the theoretical yield of a reaction, you need to follow these steps:
1. Write the balanced chemical equation:
Cu2S + O2 → 2Cu + SO2
2. Determine the molar mass of Cu2S:
Copper (Cu) has a molar mass of 63.55 g/mol, and sulfur (S) has a molar mass of 32.07 g/mol. Therefore, the molar mass of Cu2S is:
(2 × 63.55 g/mol) + (1 × 32.07 g/mol) = 159.17 g/mol
3. Convert the given mass of Cu2S (53.2g) into moles:
Moles = Mass (g) / Molar mass (g/mol)
Moles of Cu2S = 53.2 g / 159.17 g/mol = 0.334 moles
4. Use the stoichiometry of the balanced equation to find the moles of Cu produced:
From the balanced equation, we can see that 1 mole of Cu2S produces 2 moles of Cu. So,
Moles of Cu = 2 × Moles of Cu2S = 2 × 0.334 moles = 0.668 moles
5. Convert the moles of Cu into grams of Cu:
Grams = Moles × Molar mass
Grams of Cu = 0.668 moles × 63.55 g/mol = 42.43 grams
Therefore, the theoretical yield of Cu when 53.2 g of Cu2S reacts with oxygen is 42.43 grams.