Question

solve its class 11 question​

Answer 1

SEE THE AATACHMENT YOUR ANSWER IS THIS AND I THINK IT IS CORRECT

#TOGETHER WE GO FAR✔✔

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore V_{O_{2}}=168\:l}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline { \underline \bold{Given : }} \\ \implies n( C_{5} H_{10}) = 1 \: mole \\ \\ \underline { \underline \bold{To \: Find : }} \\ \implies V_{O_{2} } = ?$

• According to given question :

$\text{reaction : } \\ \bold{C_{5} H_{10} + \frac{15}{2} O_{2} \longrightarrow 5CO_{2} + 5H_{2}O} \\ \\ \text{Using \: mole \: mole \: analysis : } \\ \implies \frac{n( C_{5} H_{10} )}{S.C(C_{5} h_{10})} = \frac{n( O_{2} )}{S.C( O_{2}) } \\ \\ \text{n = No. \: of \: moles} \\ \text{S.C= stichiometric \: coefficient} \\ \\ \implies \frac{1}{1} = \frac{n( O_{2}) }{ \frac{15}{2} } \\ \\ \implies 15 = 2 \times n( O_{2}) \\ \\ \implies n( O_{2}) = \frac{15}{2} \\ \\ \text{For \: volume \: of } \: O_{2} \: at \: N.T.P\\ \implies V_{O _{2}} = n( O_{2}) \times 22.4 \\ \\ \implies V_{O _{2}} = \frac{15}{2} \times 22.4 \\ \\ \implies V_{O _{2}} =168 \: l$