Question

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

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

Solution

## The correct option is A $$30$$mlThe 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

Suggest Corrections

0

Similar questions
View More