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Ideal Gas Equation

Calculate the...

Question

Calculate the number of hours of service that can be derived at $1 atm$ and $300 K$ from an acetylene lamp containing $640 g$ of calcium carbide. Given that the lamp requires $50 L$ acetylene gas at $1 atm$ and $300 K$ for one hour. [Take $0.0821 \times 300 = 25$ ]

5 hr

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6 hr

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5.5 hr

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6.5 hr

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Solution

The correct option is A 5 hr
$C a C_{2} + 2 H_{2} O \to C a (O H)_{2} + C_{2} H_{2}$
$N u m b e r o f m o l e s o f C a C_{2} = \frac{G i v e n m a s s o f C a C_{2}}{M o l e c u l a r M a s s o f C a C_{2}}$
= $640 / 64 = 10 mol$
Since 1 mole of $C a C_{2}$ produces 1 mole of $C_{2} H_{2}$
So, 10 moles of $C a C_{2}$ will produce 10 moles of $C_{2} H_{2}$
According to the ideal gas equation,

$P V = n R T$

or
$V = \frac{n R T}{P} = \frac{10 \times 0.0821 \times 300}{1} = 250 L$
$N u m b e r o f h o u r s o f s e r v i c e^{'} T^{'} = \frac{T o t a l a m o u n t o f A c e t y l e n e p r o d u c e d}{A m o u n t o f a c e t y l e n e u s e d p e r h o u r}$
$T = \frac{250 L}{50 L/hr} = 5 hr$

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Similar questions

1 atm

300 K

640 g

50 L

1 atm

300 K

0.0821 \times 300 = 25

16 g

250 m L

300 K

a = 2.253 a t m L^{2} m o l^{- 2}, b = 0.0428 L m o l^{- 1}

R - 0.0821 L a t m K^{- 1} m o l^{- 1}

28 g N_{2}

16 g O_{2}

8.21 L

300 K

R = 0.0821 {L atm mol}^{- 1} K^{- 1}

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