Question

Uranium can be isolated from its Ores by dissolving it as UO2(NO3)2, then separating If as solid UO2(C2O4).3H2O. Addition of 0.4031 gram of Sodium Oxalate, NaC2O4,to a solution containing 1.481 gram of uranyl nitrate,UO2(NO2)2, yields 1.073 gram of solid UO2(C2O4).3H2O
Na2C2O4 + UO2(NO3)2 3H2O ? UO2(C2O4).3H2O + 2NaNO3
determine the limiting reactant and the percent yield of this reaction

Answer 1

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 2

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%