Answer:
CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓
Explanation:
We identify the reactants:
CuBr₂ and Pb(CH₃COO)₂
The products will be: Cu(CH₃COO)₂ and PbBr₂
You may know these information:
Salts from acetate are soluble.
Bromide can make solid salts with these cations: Ag⁺, Pb²⁺, Hg₂²⁺, Cu⁺
PbBr₂ is formed, so this will be our precipitate
The equation is:
CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓
c well actually I remember being tought this but I cant remember it 100
Answer:
Percentage yield = 85.2%
Explanation:
Given data:
Mass of Mg = 21.3 g
Actual yield of MgO = 30.2 g
Percentage yield = ?
Solution:
Chemical equation:
2Mg + O₂ → 2MgO
Number of moles of Mg = mass/molar mass
Number of moles of Mg = 21.3 g / 24.3 g/mol
Number of moles of Mg = 0.88 mol
Now we will compare the moles of MgO with Mg.
Mg : MgO
2 : 2
0.88 : 0.88
Mass of MgO:
Mass of MgO= moles × molar mass
Mass of MgO= 0.88 mol × 40.3g/mol
Mass of MgO = 35.46 g
Actual yield of MgO = 30.2 g
Percentage yield:
Percentage yield = Actual yield/theoretical yield × 100
Percentage yield = 30.2 g/ 35.46 g × 100
Percentage yield = 85.2%
