Some Basic Concepts of Chemistry - Result Question 33

####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:

  1. 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 $\frac{207}{32} \times 1000=6468.75 g$

$$ =6.46875 kg \approx 6.47 kg $$



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