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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?

Respuesta :

Answer : The limiting reactant is, [tex]Na_2C_2O_4[/tex]

The percent yield of the reaction is, 89.22 %

Solution : Given,

Mass of [tex]Na_2C_2O_4[/tex] = 0.4031 g

Mass of [tex]UO_2(NO_3)_2[/tex] = 1.481 g

Molar mass of [tex]Na_2C_2O_4[/tex] = 134 g/mole

Molar mass of [tex]UO_2(NO_3)_2[/tex] = 394 g/mole

Molar mass of [tex]UO_2(C_2O_4).3H_2O[/tex] = 552 g/mole

First we have to calculate the moles of [tex]Na_2C_2O_4[/tex] and [tex]UO_2(NO_3)_2[/tex].

[tex]\text{ Moles of }Na_2C_2O_4=\frac{\text{ Mass of }Na_2C_2O_4}{\text{ Molar mass of }Na_2C_2O_4}=\frac{0.4031g}{134g/mole}=0.003008moles[/tex]

[tex]\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}=\frac{1.481g}{394g/mole}=0.003759moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]Na_2C_2O_4+UO_2(NO_3)_2+3H_2O\rightarrow UO_2(C_2O_4).3H_2O+2NaNO_3[/tex]

From the balanced reaction we conclude that

As, 1 mole of [tex]Na_2C_2O_4[/tex] react with 1 mole of [tex]UO_2(NO_3)_2[/tex]

So, 0.003008 moles of [tex]Na_2C_2O_4[/tex] react with 0.003008 moles of [tex]UO_2(NO_3)_2[/tex]

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

Now we have to calculate the moles of [tex]UO_2(C_2O_4).3H_2O[/tex]

From the reaction, we conclude that

As, 1 mole of [tex]Na_2C_2O_4[/tex] react to give 1 mole of [tex]UO_2(C_2O_4).3H_2O[/tex]

So, 0.003008 moles of [tex]Na_2C_2O_4[/tex] react to give 0.003008 moles of [tex]UO_2(C_2O_4).3H_2O[/tex]

Now we have to calculate the mass of [tex]UO_2(C_2O_4).3H_2O[/tex]

[tex]\text{ Mass of }UO_2(C_2O_4).3H_2O=\text{ Moles of }UO_2(C_2O_4).3H_2O\times \text{ Molar mass of }UO_2(C_2O_4).3H_2O[/tex]

[tex]\text{ Mass of }UO_2(C_2O_4).3H_2O=(0.003008moles)\times (552g/mole)=1.660g[/tex]

Theoretical yield of [tex]UO_2(C_2O_4).3H_2O[/tex] = 1.660 g

Experimental yield of [tex]UO_2(C_2O_4).3H_2O[/tex] = 1.481 g

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

[tex]\% \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}\times 100[/tex]

[tex]\% \text{ yield of the reaction}=\frac{1.481g}{1.660g}\times 100=89.22\%[/tex]

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

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