Q162CP - Chemistry: Molecular Nature Of Matter And Change

Question

High-temperature superconducting oxides hold great promise in the utility, transportation, and computer industries. (a) One superconductor is La_2-xSr_xCuO₄. Calculate the molar masses of this oxide when x = 0, x = 1, and x = 0.163. (b) Another common superconducting oxide is made by heating a mixture of barium carbonate, copper(II) oxide, and yttrium(III) oxide, followed by further heating in O2:

$$\begin{gathered} \mathbf{4}{\mathbf{BaCO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{6}\mathbf{CuO}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{Y}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{YBa}}_{\mathbf{2}}{\mathbf{Cu}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{6}\mathbf{.}\mathbf{5}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)} \\ \mathbf{2}{\mathbf{Ba}}_{\mathbf{2}}{\mathbf{Cu}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{6}\mathbf{.}\mathbf{5}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{YBa}}_{\mathbf{2}}{\mathbf{Cu}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{7}}\mathbf{\left(}\mathbf{s}\mathbf{\right)} \end{gathered}$$

When equal masses of the three reactants are heated, which reactant is limiting? (c) After the product in part (b) is removed, what is the mass percent of each reactant in the remaining solid mixture?

Step-by-Step Solution

Answer

1. The molar masses can be found.
2. The limiting reactant is BaCO₃.
3. The mass percent of each reactant in the remaining solid mixture can be found.

Step 1: Calculating the molar masses

The molar masses for x=0,

$$\begin{gathered} M_{W}ofL{a}_{2}Cu{O}_4=(2\times M_{w}ofLa)+M_{w}ofCu+\left(4\times M_{w}ofO\right) \\ =\left(2\times 138.9\frac{g}{mol}\right)+63.55\frac{g}{mol}+\left(4\times 16\frac{g}{mol}\right) \\ {M}_{w}ofL{a}_{2}Cu{O}_4=405.35\frac{g}{mol} \end{gathered}$$

The molar masses for x=1,

$$\begin{gathered} M_{w}ofLaSrCu{O}_4=M_{w}ofLa+M_{w}ofCu+(4\times M_{w}ofO) \\ =138.9\frac{g}{mol}+87.62\frac{g}{mol}+63.55\frac{g}{mol}+\left(4\times 16\frac{g}{mol}\right) \\ {M}_{w}ofLaSrCu{O}_4=354.07\frac{g}{mol} \end{gathered}$$

The molar masses for x=0.163,

$$\begin{gathered} M_{w}of L{a}_{1.837}S{r}_{0.163}Cu{O}_4=(1.837\times M_{w}ofLa)+(0.163\times M_{w}ofSr)+M_{w}ofCu+(4\times M_{w}ofO) \\ =\left(1.837\times 138.9\frac{g}{mol}\right)+\left(87.62\frac{g}{mol}\right)+63.55\frac{g}{mol}+\left(4\times 16\frac{g}{mol}\right) \\ {M}_{w}ofLaSrCu{O}_4=396.99\frac{g}{mol} \end{gathered}$$

Step 2: Finding the limiting reactant

The moles of product can be denoted as,

$$\begin{gathered} Moles of product\left(fromBaC{O}_3\right)=xgBaC{O}_3\times \frac{1molBaC{O}_3}{197.34gBaC{O}_3}\times \frac{2molYB{a}_{2}C{u}_3{O}_{6.5}}{4molBaC{O}_3} \\ Moles of product\left(fromBaC{O}_3\right)=2.53\times {10}^{3}xmol \\ Moles of product\left(from{Y}_2{O}_3\right)=xg{Y}_2{O}_3\times \frac{1mol{Y}_2{O}_3}{225.81g{Y}_2{O}_3}\times \frac{2molYB{a}_{2}C{u}_3{O}_{6.5}}{1mol{Y}_2{O}_3}=8.86\times {10}^{-3}xmol \end{gathered}$$

The limiting reactant is BaCO₃.

Step 3: Calculating the mass percent of each reactant in remaining solid mixture

On calculating the mass percent,

$$\begin{gathered} Mass of CuOreacted=2.53\times {10}^{-3}x molproduct\times \frac{6molCuO}{2molproduct}\times \frac{79.55CuO}{1molCuO} \\ =0.6038x gCuO \\ Mass of excessCuO=xg-0.6038g=0.3962xgCuO \\ Mass of excess{Y}_2{O}_3=xg-0.2856xg=0.7144g{Y}_2{O}_3 \end{gathered}$$

Then dividing,

$$\begin{gathered} Mass\%ofexcessCuO=\frac{0.3962xg}{x}\times 100=39.62\%CuOremaining \\ Mass\%ofexcess{Y}_2{O}_3=\frac{0.7144xg}{x}\times 100=71.44\%Y_2{O}_{3}remaining \end{gathered}$$