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 > uo2(c2o4)*3h2o + 2nano3
1. determine the limiting reactant and the percent yield of this reaction?

Answer : The limiting reactant is,

The percent yield of the reaction is, 89.22 %

Solution : Given,

Mass of = 0.4031 g

Mass of = 1.481 g

Molar mass of = 134 g/mole

Molar mass of = 394 g/mole

Molar mass of = 552 g/mole

First we have to calculate the moles of and .

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 1 mole of react with 1 mole of

So, 0.003008 moles of react with 0.003008 moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of

From the reaction, we conclude that

As, 1 mole of react to give 1 mole of

So, 0.003008 moles of react to give 0.003008 moles of

Now we have to calculate the mass of

Theoretical yield of = 1.660 g

Experimental yield of = 1.481 g

Now we have to calculate the percent yield of the reaction.

Therefore, the percent yield of the reaction is, 89.22 %

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