You visited us 2 times! Enjoying our articles? Unlock Full Access!

Heat of Formation

When 12.0 g o...

Question

When 12.0 g of carbon reacted with oxygen to form CO and $C O_{2}$ at $25^{\circ} C$ and constant pressure, 75.0 kcal of heat was liberated and no carbon remained. Calculate the mass of oxygen needed for it and the mole of CO and $C O_{2}$ formed.

[Given that : $Δ_{f} H C O_{2} = - 94.05$ and $Δ_{f} H C O = - 26.41 k c a l m o l^{- 1}]$

27.49 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

32 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

17.49 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $27.49 g$
Given, $C + O_{2} ⟶ C O_{2}; Δ H = - 94.05 k c a l$
$C + \frac{1}{2} O_{2} ⟶ C O; Δ H = - 26.41 k c a l$

Let $a$ mole of carbon reacts to form $C O_{2}$ and $b$ mole of carbon reacts to form CO.

Since, only 12 g carbon; $i . e .,$ 1 mole is used up, thus,

$a + b = \frac{12}{12} = 1$ ...(i)

Also, $a \times 94.05 + b \times 26.41 = 75$ ...(ii)

$∴ a = 0.718, i . e ., m o l e o f C O_{2} f o r m e d$

$b = 0.282, i . e ., m o l e o f C O f o r m e d$

$A l s o, m o l e o f O_{2} u s e d f o r t h i s c h a n g e = 0.718 + \frac{0.282}{2} = 0.859$

$T h u s, m a s s o f O_{2} u s e d = 0.859 \times 32 = 27.49 g$

Suggest Corrections

Similar questions

C O

C O_{2}

25^{0} C

77.5

Δ_{f} H (C O) = - 25

Δ_{f} H (C O_{2}) = - 95

C O

C O_{2}

25^{\circ} C

Δ H_{f} (C O_{2}) = - 94 k c a l / m o l

Δ H_{f} (C O) = - 26.5 k c a l / m o l

C O_{2}

N_{2}

C O_{2}

27^{0} C

O_{2}

N_{2}

C O_{2}

C + O_{2} \to C O_{2}, Δ H = - 100 K c a l / m o l

C + \frac{1}{2} O_{2} \to C O, Δ H = - 25 K c a l / m o l

Join BYJU'S Learning Program