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Mole Concept and Avogadro's Number

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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

Q. 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

]

C a C_{2} + H_{2} O \to C_{2} H_{2} + C a {(O H)}_{2}

The reaction of calcium carbide and water produces acetylene gas (

C_{2} H_{2}

) and calcium hydroxide.
What volume of

C_{2} H_{2}

will form from

0.200 m o l

C a C_{2}

which reacts with an excess of water? Assume the gas will be at STP conditions.

(a) Calcium carbide is used for the artificial ripening of fruits. Actually the fruit ripens because of the heat evolved while calcium carbide reacts with moisture. During this reaction calcium hydroxide and acetylene gas is formed. If 200 cm³ of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below:
2C₂H₂_(g)+ 5O₂_(g) → 4CO₂_(g) + 2H₂O_(g)

(b) A gaseous compound of nitrogen and hydrogen contains 12.5% hydrogen by mass. Find the molecular formula of the compound if its relative molecular mass is 37.
[N = 14, H = 1].

Q. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes

2

hours on grinding/cutting machine and

3

hours on the sprayer to manufacture a pedestal lamp. It takes

1

hour on the grinding/cutting machine and

2

hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at least for at the most

20

hours and the grinding/cutting machine for at the most

12

hours. The profit from the sale of a lamp is Rs.

5

and that from a shade is Rs.

3

. Assuming that the manufacturer can sell all the lamps and shades that the produces , how should he schedule his daily production in order to maximise his profit?

Calcium carbide is used for the artificial ripening of fruits. Actually, the fruit ripens because of the heat evolved while calcium carbide reacts with moisture. During the reaction, calcium hydroxide and acetylene gas is formed.

If 200 $c m^{3}$ of acetylene is formed from a certain mass of calcium carbide. Find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below:

$2 C_{2} H_{2} (g) + 5 O_{2} (g) \to 4 C O_{2} (g) + 2 H_{2} O (g)$

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Mole Concept and Avogadro's Number

Standard X Chemistry

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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 0.0821×300=25]

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 $0.0821 \times 300 = 25$ ]