What is the volume of $14 g$ of carbon monoxide at STP?

14 L

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11.2 L

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7.1 L

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22.4 L

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Solution

The correct option is A $11.2 L$
$1 M o l e$ of gas occupy $22.4 L i t r e s$ of volume at STP.
Given, mass of $C O = 14 g$
Molar mass of $C O = 12 + 16 = 28 g$
No. of moles of $C O = \frac{14}{28} = 0.5 M o l e s$
So, $C O$ occupies $0.5 \times 22.4 = 11.2 L$ at STP.

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Similar questions

11.2 L

of gas at STP weight 14g. The gas would be :

(a) Calculate the number of moles in 7 g of nitrogen.
(b) What is the volume at STP of 7.1 g of chlorine?
(c) What is the mass of 56 $c m^{3}$ of carbon monoxide at STP?

Q. 6 g of

O_{2}

gas at STP is expanded against atmosphere so that the volume is doubled. Thus, the work done is:

^{'} X^{'}

litres of carbon monoxide is present at

S T P

. It is completely oxidized to

C O_{2}

. The volume of

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11.207

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. What is the value of

^{'} X^{'}

in litres?

Q. The volume of phosgene formed at STP when

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litres of Carbon monoxide reacts with chlorine gas is:

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What is the volume of 14g of carbon monoxide at STP?

The correct option is A 11.2L1 Mole of gas occupy 22.4 Litres of volume at STP.Given, mass of CO=14 gMolar mass of CO=12+16=28 gNo. of moles of CO=1428=0.5 MolesSo, CO occupies 0.5×22.4=11.2 L at STP.

What is the volume of $14 g$ of carbon monoxide at STP?

The correct option is A $11.2 L$
$1 M o l e$ of gas occupy $22.4 L i t r e s$ of volume at STP.
Given, mass of $C O = 14 g$
Molar mass of $C O = 12 + 16 = 28 g$
No. of moles of $C O = \frac{14}{28} = 0.5 M o l e s$
So, $C O$ occupies $0.5 \times 22.4 = 11.2 L$ at STP.