Question

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 1

Answer : The limiting reactant is,
`Na_2C_2O_4`

The percent yield of the reaction is, 89.22 %

Solution : Given,

Mass of
`Na_2C_2O_4` = 0.4031 g

Mass of
`UO_2(NO_3)_2` = 1.481 g

Molar mass of
`Na_2C_2O_4` = 134 g/mole

Molar mass of
`UO_2(NO_3)_2` = 394 g/mole

Molar mass of
`UO_2(C_2O_4).3H_2O` = 552 g/mole

First we have to calculate the moles of
`Na_2C_2O_4` and
`UO_2(NO_3)_2`.

`\text{ Moles of }Na_2C_2O_4=\frac{\text{ Mass of }Na_2C_2O_4}{\text{ Molar mass of }Na_2C_2O_4}=(0.4031g)/(134g/mole)=0.003008moles`

`\text{ Moles of }UO_2(NO_3)_2=\frac{\text{ Mass of }UO_2(NO_3)_2}{\text{ Molar mass of }UO_2(NO_3)_2}=(1.481g)/(394g/mole)=0.003759moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`Na_2C_2O_4+UO_2(NO_3)_2+3H_2O\rightarrow UO_2(C_2O_4).3H_2O+2NaNO_3`

From the balanced reaction we conclude that

As, 1 mole of
`Na_2C_2O_4` react with 1 mole of
`UO_2(NO_3)_2`

So, 0.003008 moles of
`Na_2C_2O_4` react with 0.003008 moles of
`UO_2(NO_3)_2`

From this we conclude that,
`UO_2(NO_3)_2` is an excess reagent because the given moles are greater than the required moles and
`Na_2C_2O_4` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`UO_2(C_2O_4).3H_2O`

From the reaction, we conclude that

As, 1 mole of
`Na_2C_2O_4` react to give 1 mole of
`UO_2(C_2O_4).3H_2O`

So, 0.003008 moles of
`Na_2C_2O_4` react to give 0.003008 moles of
`UO_2(C_2O_4).3H_2O`

Now we have to calculate the mass of
`UO_2(C_2O_4).3H_2O`

`\text{ Mass of }UO_2(C_2O_4).3H_2O=\text{ Moles of }UO_2(C_2O_4).3H_2O* \text{ Molar mass of }UO_2(C_2O_4).3H_2O`

`\text{ Mass of }UO_2(C_2O_4).3H_2O=(0.003008moles)* (552g/mole)=1.660g`

Theoretical yield of
`UO_2(C_2O_4).3H_2O` = 1.660 g

Experimental yield of
`UO_2(C_2O_4).3H_2O` = 1.481 g

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

`\% \text{ yield of the reaction}=\frac{\text{ Experimental yield of }UO_2(C_2O_4).3H_2O}{\text{ Theretical yield of }UO_2(C_2O_4).3H_2O}* 100`

`\% \text{ yield of the reaction}=(1.481g)/(1.660g)* 100=89.22\%`

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