Answer
$HCl$ will be used first, producing 23.4 g of $Cl_2$.
Work Step by Step
- Calculate or find the molar mass for $ HCl $:
$ HCl $ : ( 35.45 $\times$ 1 )+ ( 1.008 $\times$ 1 )= 36.46 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 48.2 \space g \times \frac{1 \space mole}{ 36.46 \space g} = 1.32 \space moles$$
Find the amount of product if each reactant is completely consumed.
$$ 0.86 \space mole \space MnO_2 \times \frac{ 1 \space mole \ Cl_2 }{ 1 \space mole \space MnO_2 } = 0.86 \space mole \space Cl_2 $$
$$ 1.32 \space moles \space HCl \times \frac{ 1 \space mole \ Cl_2 }{ 4 \space moles \space HCl } = 0.330 \space mole \space Cl_2 $$
Since the reaction of $ HCl $ produces less $ Cl_2 $ for these quantities, it is the limiting reactant.
Therefore, the $HCl$ will be used first, producing 0.330 mole of $Cl_2$.
$ Cl_2 $ : ( 35.45 $\times$ 2 )= 70.90 g/mol
$$ 0.330 \space mole \times \frac{ 70.90 \space g}{1 \space mole} = 23.4 \space g$$
