Question #d16b6

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Question #d16b6

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Answer 1 · 2017-06-06T04:04:40

$81.44 %$ $81.44%$

Explanation:

Mass of $A l {(N O_{3})}_{2}$ $Al(NO_3)_2$ - $2.5 k g$ $2.5kg$
Mass of $M {g (N O_{3})}_{2}$ $Mg(NO_3)_2$ - $2.0 k g$ $2.0kg$

-Conversion Factors-

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

$A l {(N O_{3})}_{2}$ $Al(NO_3)_2$ molar mass
$= 26.98 \frac{g}{m o l} + 2 (14.01 \frac{g}{m o l} + 3 \cdot 16.00 \frac{g}{m o l})$ $=26.98g/(mol)+2(14.01g/(mol)+3*16.00g/(mol))$
$= 151 \frac{g}{m o l}$ $=151g/(mol)$

$M {g (N O_{3})}_{2}$ $Mg(NO_3)_2$ molar mass
$= 24.31 \frac{g}{m o l} + 2 (14.01 \frac{g}{m o l} + 3 \cdot 16.00 \frac{g}{m o l})$ $=24.31g/(mol)+2(14.01g/(mol)+3*16.00g/(mol))$
$= 148.33 \frac{g}{m o l}$ $=148.33g/(mol)$

Conversion Equation
$A l {(N O_{3})}_{2 m a s s} \cdot \frac{A l {(N O_{3})}_{2 m o l s}}{A l {(N O_{3})}_{2 m a s s}} \cdot \frac{M {g (N O_{3})}_{2 m o l s}}{A l {(N O_{3})}_{2 m o l s}} \cdot \frac{M {g (N O_{3})}_{2 m a s s}}{M {g (N O_{3})}_{2 m o l s}}$ $Al(NO_3)_(2mass)*(Al(NO_3)_(2mols))/(Al(NO_3)_(2mass))*(Mg(NO_3)_(2mols))/(Al(NO_3)_(2mols))*(Mg(NO_3)_(2mass))/(Mg(NO_3)_(2mols))$
$= 2.5 k g A l {(N O_{3})}_{2} \cdot \frac{1 m o l A l {(N O_{3})}_{2}}{151 g A l {(N O_{3})}_{2}} \cdot \frac{1 m o l M {g (N O_{3})}_{2}}{1 m o l A l {(N O_{3})}_{2}} \cdot \frac{148.33 g M {g (N O_{3})}_{2}}{1 m o l M {g (N O_{3})}_{2}}$ $=2.5kgAl(NO_3)_2*(1molAl(NO_3)_2)/(151gAl(NO_3)_2)*(1molMg(NO_3)_2)/(1molAl(NO_3)_2)*(148.33gMg(NO_3)_2)/(1molMg(NO_3)_2)$
$= 2455.79 g M {g (N O_{3})}_{2}$ $=2455.79gMg(NO_3)_2$

Percent Yield
$%Yield = \frac{Actual}{Theoretical}$ $"%Yield"=("Actual")/("Theoretical")$
$= \frac{2000 g M {g (N O_{3})}_{2}}{2455.79 g M {g (N O_{3})}_{2}} = .8144 = 81.44 %$ $=(2000gMg(NO_3)_2)/(2455.79gMg(NO_3)_2)=.8144=81.44%$

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Question #d16b6

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