Q73E - Chemistry (OpenStax)

Question

Uranium can be isolated from its ores by dissolving it as UO₂(NO₃)₂, then separating it as solid UO₂(C₂O₄). Addition of 0.4031 g of sodium oxalate, NaC₂O₄, to a solution containing 1.481 g of uranyl nitrate, UO₂(NO₃)₂, yields 1.073 g of solid

\(Na{C_2}{O_4} + U{O_2}{\left( {N{O_3}} \right)_2} + 3{H_2}O \to U{O_2}\left( {{C_2}{O_4}} \right) \cdot 3{H_2}O + 2NaN{O_3}\)

Step-by-Step Solution

Answer

NaC₂O₄ is the limiting reactant.

The percent yield is 86.6 %.

Balance chemical equation

Balance the chemical equation

Determine the molar mass

Find the molar mass.

1 mol                    =2(22.990) +2(12.011) +4(15.999) =133.998 g.

1mol    = 238.029+8(15.999)+2(14.007) = 394.035 g.

1 mol    = 238.029+9(15.999) +2(12.011) +6(1.008) = 412.09 g

Determine the yield

Find the theoretical yield 

\(\begin{aligned}{\underline{\phantom{xx}}}3.0083 \times {10^{ - 3}}\,mol\,Na{C_2}{O_4} \times \left( {\frac{{1\,mol\,U{O_2}\left( {{C_2}{O_4}} \right) \cdot 3{H_2}O}}{{1\,mol\,mol\,Na{C_2}{O_4}}}} \right) \times \\\left( {\frac{{412.09\,g\,U{O_2}\left( {{C_2}{O_4}} \right) \cdot 3{H_2}O}}{{1\,mol\,U{O_2}\left( {{C_2}{O_4}} \right) \cdot 3{H_2}O}}} \right)\\ = 1\,.239\,g\,U{O_2}\left( {{C_2}{O_4}} \right) \cdot 3{H_2}O\end{aligned}\)

Find the percent yield of