## Chemistry 10th Edition

(a) The mixture of reactants will have 8 CO molecules and 6 $O_2$ molecules. (b) The product mixture has 2 $O_2$ molecules and 8 $CO_2$ molecules. (c) 211.6 g of $CO_2$
(a) The reactants mixture is made of 8 CO molecules and 6 $O_2$ molecules, as said in the exercise. (b) 8 CO molecules can produce 8 $CO_2$ molecules. 6 $O_2$ molecules can produce 12 $CO_2$ molecules. Thus, $CO$ is the limiting reactant, and 8 $CO_2$ molecules will be produced. To produce 8 $CO_2$ molecules, we need 4 $O_2$. So, 2 oxygen molecules will remain in the products. (c) - 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: $$134.67 \space g \times \frac{1 \space mole}{ 28.01 \space g} = 4.808 \space moles$$ - Calculate or find the molar mass for $O_2$: $O_2$ : ( 16.00 $\times$ 2 )= 32.00 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$77.25 \space g \times \frac{1 \space mole}{ 32.00 \space g} = 2.414 \space moles$$ Find the amount of product if each reactant is completely consumed. $$4.808 \space moles \space CO \times \frac{ 2 \space moles \ CO_2 }{ 2 \space moles \space CO } = 4.808 \space moles \space CO_2$$ $$2.414 \space moles \space O_2 \times \frac{ 2 \space moles \ CO_2 }{ 1 \space mole \space O_2 } = 4.828 \space moles \space CO_2$$ Since the reaction of $CO$ produces less $CO_2$ for these quantities, it is the limiting reactant. $CO_2$ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol - Using the molar mass as a conversion factor, find the mass in g: $$4.808 \space mole \times \frac{ 44.01 \space g}{1 \space mole} = 211.6 \space g$$