Answer
When 12.8 g of Na and 10.2 g of $Cl_2$ reacts, 16.8 g of $NaCl$ is produced.
Work Step by Step
- Calculate or find the molar mass for $ Na $:
$ Na $ : 22.99 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 12.8 \space g \times \frac{1 \space mole}{ 22.99 \space g} = 0.557 \space mole$$
- 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:
$$ 10.2 \space g \times \frac{1 \space mole}{ 70.90 \space g} = 0.144 \space mole$$
Find the amount of product if each reactant is completely consumed.
$$ 0.557 \space mole \space Na \times \frac{ 2 \space moles \ NaCl }{ 2 \space moles \space Na } = 0.557 \space mole \space NaCl $$
$$ 0.144 \space mole \space Cl_2 \times \frac{ 2 \space moles \ NaCl }{ 1 \space mole \space Cl_2 } = 0.288 \space mole \space NaCl $$
Since the reaction of $ Cl_2 $ produces less $ NaCl $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for $ NaCl $:
$ NaCl $ : ( 22.99 $\times$ 1 )+ ( 35.45 $\times$ 1 )= 58.44 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.288 \space mole \times \frac{ 58.44 \space g}{1 \space mole} = 16.8 \space g$$
