According to equation, C6H6(l)+(15/2)O2(g) xrightarrow3H2O(l)+6CO2(g) Δ textH=-3264.4 text kJ/mol . The energy evolved when 7.8 g of benzene is burnt in air will be:

Question

According to equation, C6H6(l)+(15/2)O2(g) xrightarrow3H2O(l)+6CO2(g) Δ textH=-3264.4 text kJ/mol . The energy evolved when 7.8 g of benzene is burnt in air will be: (A)  3.264 text kJ/mol  (B)  32.64 text kJ/mol  (C)  326.4 text kJ/mol  (D)  163.22 text kJ/mol

Tardigrade Admin · Accepted Answer

underset(6× 12+6× 1=78) mathopC6H6(l)+(15/2)O2(g) → 3H2O(l)+6CO2(g) ∵ The energy evolved when 78 g. benzene is burnt = 3264.4 kJ/mol ∴ The energy evolved when 7.8g benzene is burnt =(3264.4× 7.8/78)=326.44kJ/mol

Q. According to equation, $C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) 3 H_{2} O (l) + 6 C O_{2} (g)$ $Δ H = - 3264.4 k J / m o l$ . The energy evolved when 7.8 g of benzene is burnt in air will be:

$3.264 k J / m o l$

$32.64 k J / m o l$

$326.4 k J / m o l$

$163.22 k J / m o l$

$(6 \times 12 + 6 \times 1 = 78) C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 3 H_{2} O (l) + 6 C O_{2} (g)$ $∵$ The energy evolved when 78 g. benzene is burnt = 3264.4 kJ/mol $∴$ The energy evolved when 7.8g benzene is burnt $= \frac{3264.4 \times 7.8}{78} = 326.44 k J / m o l$

Q. According to equation, C6​H6​(l)+215​O2​(g)​3H2​O(l)+6CO2​(g) ΔH=−3264.4kJ/mol . The energy evolved when 7.8 g of benzene is burnt in air will be:

3.264kJ/mol

32.64kJ/mol

326.4kJ/mol

163.22kJ/mol