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Oxidation Number Method

What mass of ...

Question

What mass of carbon disulphide, $C S_{2}$ can be completely oxidized to $S O_{2}$ and $C O_{2}$ by the oxygen liberated when $325 g$ of $N a_{2} O_{2}$ react with water?

316.67 g

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52.78 g

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633.33 g

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211.11 g

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Solution

The correct option is B $52.78 g$
Given,
$325 g$ of $N a_{2} O_{2}$ is reacted with water
Molar mass of $N a_{2} O_{2} = 78 g / m o l$
$∴$ Number of moles of $N a_{2} O_{2}$ given= $\frac{325 g}{78 g / m o l}$
$= 4.166$ moles
The reaction involved is
$2 N a_{2} O_{2} + 2 H_{2} O ⟶ 4 N a O H + O_{2}$
$2$ moles of $N a_{2} O_{2}$ gives $1$ moles of $O_{2}$
$∴ 4.166$ moles of $N a_{2} O_{2}$ gives $⟶ \frac{4.166}{2}$ moles of $O_{2}$
$= 2.083$ moles of $O_{2}$
Now,
The molar mass of $C S_{2} = 76 / g m o l$
The reaction involved id
$C S_{2} + 3 O_{2} ⟶ C O_{2} + 2 S O_{2}$
$1$ mole of $C S_{2}$ requires $3$ moles of $O_{2}$
Now, $2.083$ moles of $O_{2}$ will react with $x$ moles of $C S_{2}$ to produce $S O_{2}$ & $C O_{2}$
$x = 0.6943$ moles of $C S_{2}$
$∴$ The mass of $C S_{2}$ which can be completely oxidized by $2.083$ moles of $O_{2} = 0.6943 m o l \times 76 g / m o l$
$= 52.76 g$

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What mass of carbon disulphide, CS2 can be completely oxidized to SO2 and CO2 by the oxygen liberated when 325 g of Na2O2 react with water?

What mass of carbon disulphide, $C S_{2}$ can be completely oxidized to $S O_{2}$ and $C O_{2}$ by the oxygen liberated when $325 g$ of $N a_{2} O_{2}$ react with water?