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$${ C }_{ 6 }{ H }_{ 5 }OH(g)+O_{ 2 }(g)\longrightarrow { CO }_{ 2 }(g)+{ H }_{ 2 }O(l)$$
Magnitude of volume change if $$30$$ml of $${ C }_{ 6 }{ H }_{ 5 }OH(g)$$ is burnt with excess amount of oxygen, is:


A
30ml
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B
60ml
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C
20ml
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D
10ml
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Solution

The correct option is A $$30$$ml

The given equation is

$${{C}_{6}}{{H}_{5}}O{{H}_{\left( g \right)}}+{{O}_{2\left( g \right)}}\xrightarrow{{}}C{{O}_{2\left( g \right)}}+{{H}_{2}}{{O}_{\left( l \right)}}$$

After balanced equation is

$${{C}_{6}}{{H}_{5}}O{{H}_{\left( g \right)}}+7{{O}_{2\left( g \right)}}\xrightarrow{{}}6C{{O}_{2\left( g \right)}}+3{{H}_{2}}{{O}_{\left( l \right)}}$$


Volume of reactant is;

Number of moles of $${{C}_{6}}{{H}_{5}}OH=1$$

Volume of $${{C}_{6}}{{H}_{5}}OH=30\ mL$$

$$\therefore $$ Moles of $$Oxygen=7$$

Volume $$=7\times 30=210\ mL$$


Volume of products is ;

Moles of $$C{{O}_{2}}=6$$

Volume $$=6\times 30=180\ mL$$

Moles of $${{H}_{2}}O=3$$

Volume $$=3\times 30=90\ mL$$

therefore,

$$\therefore $$ Change in volume$$\ =\Pr oduct\ volume\ -\ reac\tan t\ volume$$


  $$ \Rightarrow \left( 180+90 \right)-\left( 210+30 \right) $$

 $$ \Rightarrow 270-240 $$

 $$ \Rightarrow 30\ mL $$


Hence;

This is the required solution.

Chemistry

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