## Chemistry: Molecular Approach (4th Edition)

2.91 g of $CO$ remain in the mixture.
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: $$0.104 \space mole \times \frac{ 28.01 \space g}{1 \space mole} = 2.91 \space g$$