What is the maximum amount of $B a {S O}_{4}$ in terms of moles precipitated on mixing equal volumes $1$ liter each of $B a {C l}_{2}$ $2.05 M$ with $H_{2} {S O}_{4}$ $1 M$ ?

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Solution

Given,
Molarity of $B a C l_{2} = 2.05 M$
Volume of $B a C l_{2} = 1 L$
Molarity of $H_{2} S O_{4} = 1 M$
Volume of $H_{2} S O_{4} = 1 L$
$\Rightarrow$ Moles of $B a C l_{2} = 2.05 M \times 1 L = 2.05$ moles
Moles of $H_{2} S O_{4} = 1 M \times 1 L = 1$ mole
The reaction involved is
$B a C l_{2} + H_{2} S O_{4} ⟶ B a S O_{4} + 2 H C l$
It is clear that, $1$ mole $B a C l_{2}$ consume $1$ mole $H_{2} S O_{4}$ to produce $1$ mole $B a S O_{4}$ .
$\Rightarrow 2.05$ moles of $B a C l_{2}$ consume $2.05$ moles of $H_{2} S O_{4}$ to produce $2.05$ moles $B a S O_{4}$ but only $1$ mole of $H_{2} S O_{4}$ is provided
$∴ H_{2} S O_{4}$ here is limiting reagent.
$\Rightarrow 1$ mole $H_{2} S O_{4}$ consumes $1$ mole $B a C l_{2}$ and produces $1$ mole $B a S O_{4}$ according to above reaction.
$∴ 1$ mole $B a S O_{4}$ is maximum produced here.

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