Answer:
The limiting reactant is NaC₂O₄ and the yield of this reaction is 69.52%.
Explanation:
NaC₂O₄ + UO₂(NO₃)₂ + 3H₂O → UO₂(C₂O₄).3H₂O + 2NaNO₃
m (NaC₂O₄) = 0.4031 g MW = 134 g/mol ∴ n = 3.0 mmol
m (UO₂(NO₃)₂) = 1.481 g MW = 396.05 g/mol ∴ n = 3.74 mmol
m (UO₂(C₂O₄).3H₂O) = 1.073 g MW = 412.094 g/mol ∴ n = 2.60 mmol
The limiting reactant is NaC₂O₄
The yield can be given by (2.60/3.74).100% = 69.52%
Answer:
Limiting reactant: Na₂C₂O₄
Percent yield: 86.53%
Explanation:
Let's consider the following reaction.
Na₂C₂O₄ + UO₂(NO₃)₂ + 3H₂O → UO₂(C₂O₄).3H₂O + 2NaNO₃
The molar mass of Na₂C₂O₄ is 134.0 g/mol and the molar mass of UO₂(NO₃)₂ is 394.0 g/mol. In the balanced equation, there is 1 mole of each one.
The theoretical mass ratio of UO₂(NO₃)₂ to Na₂C₂O₄ is 394.0/134.0 = 2.940/1.
The experimental mass ratio of UO₂(NO₃)₂ to Na₂C₂O₄ is 1.481/0.4031 = 3.674/1.
Comparing the theoretical and the experimental mass ratios, we can see that the reactant in excess is UO₂(NO₃)₂ and the limiting reactant is Na₂C₂O₄.
We will use the limiting reactant to calculate the theoretical yield of UO₂(C₂O₄).3H₂O. The molar mass of UO₂(C₂O₄).3H₂O is 412.1 g/mol and the mass ratio of Na₂C₂O₄ to UO₂(C₂O₄).3H₂O is 134.0g / 412.1g.
The theoretical yield of UO₂(C₂O₄).3H₂O is:
0.4031 g Na₂C₂O₄ × (412.1 g UO₂(C₂O₄).3H₂O / 134.0 g Na₂C₂O₄) = 1.240 g UO₂(C₂O₄).3H₂O
The percent yield is:
(1.073 g / 1.240 g) × 100% = 86.53%
