Some Basic Concepts of Chemistry - Result Question 95
####33. How many millilitres of $0.5 M H _2 SO _4$ are needed to dissolve $0.5 g$ of copper (II) carbonate?
(1999, 3M)
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Solution:
- The equations of chemical reactions occurring during the process are
In the presence of oxygen
$$ 2 PbS+3 O _2 \longrightarrow 2 PbO+2 SO _2 $$
By self reduction
$$ 2 PbO+PbS \longrightarrow 3 Pb+SO _2 $$
Thus 3 moles of $O _2$ produces 3 moles of $Pb$
i.e. $32 \times 3=96 g$ of $O _2$ produces $3 \times 207=621 g$ of $Pb$
So $1000 g(1 kg)$ of oxygen will produce
$$ \begin{aligned} \frac{621}{96} \times 1000 & =6468.75 g \ & =6.4687 kg \approx 6.47 kg \end{aligned} $$
Alternative Method
From the direct equation,
$$ PbS+\underset{32 g}{O _2} \longrightarrow \underset{207 g}{Pb}+SO _2 $$
So, $32 g$ of $O _2$ gives $207 g$ of $Pb$
$1 g$ of $O _2$ will give $\frac{207}{32} g$ of $Pb$
$1000 g$ of $ _2$ will give $\frac{207}{32} \times 1000=6468.75 g$
$$ =6.46875 kg \approx 6.47 kg $$
