Q162CP High-temperature superconducting... [FREE SOLUTION]

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Chapter 3: Q162CP (page 139)

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:
$4 {BaCO}_{3} (s) + 6 CuO (s) + Y_{2} O_{3} (s) 鈫� 2 {YBa}_{2} {Cu}_{3} O_{6.5} (s) + 4 {CO}_{2} (g) 2 {Ba}_{2} {Cu}_{3} O_{6.5} (s) + \frac{1}{2} O_{2} (g) 鈫� 2 {YBa}_{2} {Cu}_{3} O_{7} (s)$
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?

Short Answer

Expert verified

The molar masses can be found.
The limiting reactant isBaCO₃.
The mass percent of each reactant in the remaining solid mixture can be found.

Step by step solution

Calculating the molar masses

The molar masses for x=0,

$M_{W} o f L a_{2} C u O_{4} = (2 脳 M_{w} o f L a) + M_{w} o f C u + (4 脳 M_{w} o f O) = (2 脳 138.9 \frac{g}{m o l}) + 63.55 \frac{g}{m o l} + (4 脳 16 \frac{g}{m o l}) M_{w} o f L a_{2} C u O_{4} = 405.35 \frac{g}{m o l}$

The molar masses for x=1,

$M_{w} o f L a S r C u O_{4} = M_{w} o f L a + M_{w} o f C u + (4 脳 M_{w} o f O) = 138.9 \frac{g}{m o l} + 87.62 \frac{g}{m o l} + 63.55 \frac{g}{m o l} + (4 脳 16 \frac{g}{m o l}) M_{w} o f L a S r C u O_{4} = 354.07 \frac{g}{m o l}$

The molar masses for x=0.163,

$M_{w} o f L a_{1.837} S r_{0.163} C u O_{4} = (1.837 脳 M_{w} o f L a) + (0.163 脳 M_{w} o f S r) + M_{w} o f C u + (4 脳 M_{w} o f O) = (1.837 脳 138.9 \frac{g}{m o l}) + (87.62 \frac{g}{m o l}) + 63.55 \frac{g}{m o l} + (4 脳 16 \frac{g}{m o l}) M_{w} o f L a S r C u O_{4} = 396.99 \frac{g}{m o l}$

Finding the limiting reactant

The moles of product can be denoted as,

$M o l e s o f p r o d u c t (f r o m B a C O_{3}) = x g B a C O_{3} 脳 \frac{1 m o l B a C O_{3}}{197.34 g B a C O_{3}} 脳 \frac{2 m o l Y B a_{2} C u_{3} O_{6.5}}{4 m o l B a C O_{3}} M o l e s o f p r o d u c t (f r o m B a C O_{3}) = 2.53 脳 10^{3} x m o l M o l e s o f p r o d u c t (f r o m Y_{2} O_{3}) = x g Y_{2} O_{3} 脳 \frac{1 m o l Y_{2} O_{3}}{225.81 g Y_{2} O_{3}} 脳 \frac{2 m o l Y B a_{2} C u_{3} O_{6.5}}{1 m o l Y_{2} O_{3}} = 8.86 脳 10^{- 3} x m o l$

The limiting reactant is BaCO₃.

Calculating the mass percent of each reactant in remaining solid mixture

On calculating the mass percent,

$M a s s o f C u O r e a c t e d = 2.53 脳 10^{- 3} x m o l p r o d u c t 脳 \frac{6 m o l C u O}{2 m o l p r o d u c t} 脳 \frac{79.55 C u O}{1 m o l C u O} = 0.6038 x g C u O M a s s o f e x c e s s C u O = x g - 0.6038 g = 0.3962 x g C u O M a s s o f e x c e s s Y_{2} O_{3} = x g - 0.2856 x g = 0.7144 g Y_{2} O_{3}$

Then dividing,

$M a s s % o f e x c e s s C u O = \frac{0.3962 x g}{x} 脳 100 = 39.62 % C u O r e m a i n i n g M a s s % o f e x c e s s Y_{2} O_{3} = \frac{0.7144 x g}{x} 脳 100 = 71.44 % Y_{2} O_{3} r e m a i n i n g$