#### Answer

(a) Some C is left over.
(b) 23 g of C

#### Work Step by Step

1. Balance the reaction:
$$SiO_2 + 3 C \longrightarrow SiC + 2CO$$
- Calculate or find the molar mass for $ SiO_2 $:
$ SiO_2 $ : ( 28.09 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 60.09 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 300. \space g \times \frac{1 \space mole}{ 60.09 \space g} = 4.99 \space moles$$
- Calculate or find the molar mass for $ C $:
$ C $ : 12.01 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 203 \space g \times \frac{1 \space mole}{ 12.01 \space g} = 16.9 \space moles$$
Find the amount of product if each reactant is completely consumed.
$$ 4.99 \space moles \space SiO_2 \times \frac{ 1 \space mole \ SiC }{ 1 \space mole \space SiO_2 } = 4.99 \space moles \space SiC $$
$$ 16.9 \space moles \space C \times \frac{ 1 \space mole \ SiC }{ 3 \space moles \space C } = 5.63 \space moles \space SiC $$
Since the reaction of $ SiO_2 $ produces less $ SiC $ for these quantities, it is the limiting reactant.
$$ 4.99 \space moles \space SiO_2 \times \frac{ 3 \space moles \ C }{ 1 \space mole \space SiO_2 } = 15.0 \space moles \space C $$
- Using the molar mass as a conversion factor, find the mass in g:
$$ 15.0 \space mole \times \frac{ 12.01 \space g}{1 \space mole} = 180. \space g$$
Excess = Initial - Consumed = 203 g - 180. g = 23 g