Determine the limiting reactant and the percent yield Open Stax Chapter 4 number 73

Uranium can be isolated from its ores by dissolving it as UO2(NO3)2, then separating it as solid UO2(C2O4)∙3H2O. Addition of 0.4031 g of sodium oxalate, Na2C2O4, to a solution containing 1.481 g of uranyl nitrate, UO2(NO3)2, yields 1.073 g of solid UO2(C2O4)∙3H2O.

Na2C2O4 + UO2(NO3)2 + 3H2O "yields" UO2(C2O4)∙3H2O + 2NaNO3

Determine the limiting reactant and the percent yield of this reaction.

This problem comes from the open stax 1st edition chemistry textbook, chapter 4, number 73.
I take you through how to work this problem from determine the limiting reactant, calculating the theoretical yield, and then calculating the percent yield.

Step 1: Convert grams to moles using molar mass
Step 2: Convert moles to moles using the balanced chemical equation
Step 3: Convert moles to grams using molar mass
Step 4: Identify the limiting reactant and theoretical yield
Step 5: Calculate the percent yield of the reaction.

[ajt.]

Why did you use 0.4031g of sodium oxalate to figure out the grams of product and not 1.481g of uranyl nitrate? Is it because sodium oxalate was the limiting reactant? So, to find the percent yield of the product we have to use the limiting reactant from the reactant side of the equation?