Answer
Limiting reactant: $SiO_2$
Theoretical yield: 72.84 kg
Percent yield: 90.7%
Work Step by Step
- Calculate or find the molar mass for $ SiO_2 $:
$ SiO_2 $ : ( 28.09 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 60.09 kg/kmol
- Using the molar mass as a conversion factor, find the amount in kmol:
$$ 155.8 \space kg \times \frac{1 \space kmol}{ 60.09 \space kg} = 2.593 \space kmol$$
- Calculate or find the molar mass for $ C $:
$ C $ : 12.01 kg/kmol
- Using the molar mass as a conversion factor, find the amount in kmol:
$$ 78.3 \space kg \times \frac{1 \space kmol}{ 12.01 \space kg} = 6.52 \space kmol$$
Find the amount of product if each reactant is completely consumed.
$$ 2.593 \space kmol \space SiO_2 \times \frac{ 1 \space kmol \ Si }{ 1 \space kmol \space SiO_2 } = 2.593 \space kmol \space Si $$
$$ 6.52 \space kmol \space C \times \frac{ 1 \space kmol \ Si }{ 1 \space kmol \space C } = 6.52 \space kmol \space Si $$
Since the reaction of $ SiO_2 $ produces less $ Si $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for $ Si $:
$ Si $ : 28.09 kg/kmol
- Using the molar mass as a conversion factor, find the mass in kg:
$$ 2.593 \space kmol \times \frac{ 28.09 \space kg}{1 \space kmol} = 72.84 \space kg$$
$$Percent \space yield = \frac{ 66.1 }{ 72.84 } \times 100\% = 90.7 \% $$
