# Question d16b6

81.44%

#### Explanation:

Mass of $A l {\left(N {O}_{3}\right)}_{2}$- $2.5 k g$
Mass of $M g {\left(N {O}_{3}\right)}_{2}$- $2.0 k g$

-Conversion Factors-

Molar Masses:
Aluminum (Al)- $26.98 \frac{g}{m o l}$
Nitrogen(N)- $14.01 \frac{g}{m o l}$
Oxygen(O)- $16.00 \frac{g}{m o l}$
Magnesium(Mg)- $24.31 \frac{g}{m o l}$

$A l {\left(N {O}_{3}\right)}_{2}$ molar mass
$= 26.98 \frac{g}{m o l} + 2 \left(14.01 \frac{g}{m o l} + 3 \cdot 16.00 \frac{g}{m o l}\right)$
$= 151 \frac{g}{m o l}$

$M g {\left(N {O}_{3}\right)}_{2}$ molar mass
$= 24.31 \frac{g}{m o l} + 2 \left(14.01 \frac{g}{m o l} + 3 \cdot 16.00 \frac{g}{m o l}\right)$
$= 148.33 \frac{g}{m o l}$

Conversion Equation
$A l {\left(N {O}_{3}\right)}_{2 m a s s} \cdot \frac{A l {\left(N {O}_{3}\right)}_{2 m o l s}}{A l {\left(N {O}_{3}\right)}_{2 m a s s}} \cdot \frac{M g {\left(N {O}_{3}\right)}_{2 m o l s}}{A l {\left(N {O}_{3}\right)}_{2 m o l s}} \cdot \frac{M g {\left(N {O}_{3}\right)}_{2 m a s s}}{M g {\left(N {O}_{3}\right)}_{2 m o l s}}$
$= 2.5 k g A l {\left(N {O}_{3}\right)}_{2} \cdot \frac{1 m o l A l {\left(N {O}_{3}\right)}_{2}}{151 g A l {\left(N {O}_{3}\right)}_{2}} \cdot \frac{1 m o l M g {\left(N {O}_{3}\right)}_{2}}{1 m o l A l {\left(N {O}_{3}\right)}_{2}} \cdot \frac{148.33 g M g {\left(N {O}_{3}\right)}_{2}}{1 m o l M g {\left(N {O}_{3}\right)}_{2}}$
$= 2455.79 g M g {\left(N {O}_{3}\right)}_{2}$

Percent Yield
"%Yield"=("Actual")/("Theoretical")
=(2000gMg(NO_3)_2)/(2455.79gMg(NO_3)_2)=.8144=81.44%#