The enthalpy of combustion of methane at 25-C is 890 kJ- The heat liberated when 3-2 g of methane is burnt in air is-

The correct option is D 178 kJNumber of moles of methane burnt = 3.216=0.2 Enthalpy of combustion of 1 mole = 890 kJ/molHeat liberated = ...

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Standard XII

Chemistry

Enthalpy of Combustion

The enthalpy ...

Question

The enthalpy of combustion of methane at $25^{\circ} C$ is $890$ kJ. The heat liberated when $3.2$ g of methane is burnt in air is:

445

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278

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- 890

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178

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Solution

The correct option is D $178$ kJ
Number of moles of methane burnt $= \frac{3.2}{16} = 0.2$

Enthalpy of combustion of 1 mole $= 890$ kJ/mol

Heat liberated = $890 \times 0.2 = 178$ kJ

Hence, option D is correct.

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The enthalpy of combustion of methane at 25∘C is 890 kJ. The heat liberated when 3.2 g of methane is burnt in air is:

The correct option is D 178 kJNumber of moles of methane burnt =3.216=0.2Enthalpy of combustion of 1 mole =890 kJ/molHeat liberated = 890×0.2=178 kJHence, option D is correct.

The enthalpy of combustion of methane at $25^{\circ} C$ is $890$ kJ. The heat liberated when $3.2$ g of methane is burnt in air is:

The correct option is D $178$ kJ
Number of moles of methane burnt $= \frac{3.2}{16} = 0.2$

Enthalpy of combustion of 1 mole $= 890$ kJ/mol

Heat liberated = $890 \times 0.2 = 178$ kJ

Hence, option D is correct.