## Chemistry: Molecular Approach (4th Edition)

a. 2.5 g of $AlCl_3$ b. 31.1 g of $AlCl_3$ c. 1.16 g of $AlCl_3$
a. - Calculate or find the molar mass for $Al$: $Al$ : 26.98 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$2.0 \space g \times \frac{1 \space mole}{ 26.98 \space g} = 0.074 \space mole$$ - Calculate or find the molar mass for $Cl_2$: $Cl_2$ : ( 35.45 $\times$ 2 )= 70.90 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$2.0 \space g \times \frac{1 \space mole}{ 70.90 \space g} = 0.028 \space mole$$ Find the amount of product if each reactant is completely consumed. $$0.074 \space mole \space Al \times \frac{ 2 \space moles \ AlCl_3 }{ 2 \space moles \space Al } = 0.074 \space mole \space AlCl_3$$ $$0.028 \space mole \space Cl_2 \times \frac{ 2 \space moles \ AlCl_3 }{ 3 \space moles \space Cl_2 } = 0.019 \space mole \space AlCl_3$$ Since the reaction of $Cl_2$ produces less $AlCl_3$ for these quantities, it is the limiting reactant. - Calculate or find the molar mass for $AlCl_3$: $AlCl_3$ : ( 26.98 $\times$ 1 )+ ( 35.45 $\times$ 3 )= 133.33 g/mol - Using the molar mass as a conversion factor, find the mass in g: $$0.019 \space mole \times \frac{ 133.33 \space g}{1 \space mole} = 2.5 \space g$$ b. - Using the molar mass as a conversion factor, find the amount in moles: $$7.5 \space g \times \frac{1 \space mole}{ 26.98 \space g} = 0.28 \space mole$$ - Using the molar mass as a conversion factor, find the amount in moles: $$24.8 \space g \times \frac{1 \space mole}{ 70.90 \space g} = 0.350 \space mole$$ Find the amount of product if each reactant is completely consumed. $$0.28 \space mole \space Al \times \frac{ 2 \space moles \ AlCl_3 }{ 2 \space moles \space Al } = 0.28 \space mole \space AlCl_3$$ $$0.350 \space mole \space Cl_2 \times \frac{ 2 \space moles \ AlCl_3 }{ 3 \space moles \space Cl_2 } = 0.233 \space mole \space AlCl_3$$ Since the reaction of $Cl_2$ produces less $AlCl_3$ for these quantities, it is the limiting reactant. - Using the molar mass as a conversion factor, find the mass in g: $$0.233 \space mole \times \frac{ 133.33 \space g}{1 \space mole} = 31.1 \space g$$ c. - Using the molar mass as a conversion factor, find the amount in moles: $$0.235 \space g \times \frac{1 \space mole}{ 26.98 \space g} = 0.00871 \space mole$$ - Using the molar mass as a conversion factor, find the amount in moles: $$1.15 \space g \times \frac{1 \space mole}{ 70.90 \space g} = 0.0162 \space mole$$ Find the amount of product if each reactant is completely consumed. $$0.00871 \space mole \space Al \times \frac{ 2 \space moles \ AlCl_3 }{ 2 \space moles \space Al } = 0.00871 \space mole \space AlCl_3$$ $$0.0162 \space mole \space Cl_2 \times \frac{ 2 \space moles \ AlCl_3 }{ 3 \space moles \space Cl_2 } = 0.0108 \space mole \space AlCl_3$$ Since the reaction of $Al$ produces less $AlCl_3$ for these quantities, it is the limiting reactant. - Using the molar mass as a conversion factor, find the mass in g: $$0.00871 \space mole \times \frac{ 133.33 \space g}{1 \space mole} = 1.16 \space g$$