## Chemistry: Molecular Approach (4th Edition)

Limiting reactant: $Mg$ Theoretical yield: 16.7 g Percent yield: 71.3%
- Calculate or find the molar mass for $Mg$: $Mg$ : 24.31 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$10.1 \space g \times \frac{1 \space mole}{ 24.31 \space g} = 0.415 \space mole$$ - 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: $$10.5 \space g \times \frac{1 \space mole}{ 32.00 \space g} = 0.328 \space mole$$ Find the amount of product if each reactant is completely consumed. $$0.415 \space mole \space Mg \times \frac{ 2 \space moles \ MgO }{ 2 \space moles \space Mg } = 0.415 \space mole \space MgO$$ $$0.328 \space mole \space O_2 \times \frac{ 2 \space moles \ MgO }{ 1 \space mole \space O_2 } = 0.656 \space mole \space MgO$$ Since the reaction of $Mg$ produces less $MgO$ for these quantities, it is the limiting reactant. - Calculate or find the molar mass for $MgO$: $MgO$ : ( 24.31 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 40.31 g/mol - Using the molar mass as a conversion factor, find the mass in g: $$0.415 \space mole \times \frac{ 40.31 \space g}{1 \space mole} = 16.7 \space g$$ $$Percent \space yield = \frac{ 11.9 }{ 16.7 } \times 100\% = 71.3\%$$