Answer
2.91 g of $CO$ remain in the mixture.
Work Step by Step
1. Identify the limiting reactant.
- Calculate or find the molar mass for $ Fe_2O_3 $:
$ Fe_2O_3 $ : ( 55.85 $\times$ 2 )+ ( 16.00 $\times$ 3 )= 159.7 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 22.55 \space g \times \frac{1 \space mole}{ 159.7 \space g} = 0.1412 \space mole$$
- Calculate or find the molar mass for $ CO $:
$ CO $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 28.01 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 14.78 \space g \times \frac{1 \space mole}{ 28.01 \space g} = 0.528 \space mole$$
Find the amount of product if each reactant is completely consumed.
$$ 0.1412 \space mole \space Fe_2O_3 \times \frac{ 2 \space moles \ Fe }{ 1 \space mole \space Fe_2O_3 } = 0.2824 \space mole \space Fe $$
$$ 0.5277 \space mole \space CO \times \frac{ 2 \space moles \ Fe }{ 3 \space moles \space CO } = 0.3518 \space mole \space Fe $$
Since the reaction of $ Fe_2O_3 $ produces less $ Fe $ for these quantities, it is the limiting reactant.
2. Find the amount of $ CO $ consumed:
$$ 0.1412 \space moles \space Fe_2O_3 \times \frac{ 3 \space mol \space CO }{ 1 \space mol \space Fe_2O_3 } = 0.4236 \space mol \space CO $$
3. Subtract that amount from the amount in the reaction mixture.
$$Excess = 0.528 - 0.4236 = 0.104 \space mol \space CO $$
4. Using its molar mass, find the amount in grams:
$$ 0.104 \space mole \times \frac{ 28.01 \space g}{1 \space mole} = 2.91 \space g$$
