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

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%

since electrons are negative the answer is neagtaive