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You need to get the Moles of CO2 first. 9.3 g. CO2/44 g CO2= .211 mol

Then just put it into the ideal gas law at STP. PV=nRT

1.0 ATM x P = (.211 mol) (.08206) (273.15 k)---->"Plug and Chug"

P = (.211 mol) (.08206) (273.15 k) / 1.0 ATM

P= 4.7 L ---> 4,700 mL

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14y ago
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12y ago

Use the ideal gas law: PV=nRT

P= 1.00 ATM

V= 6.2 L

n= # of moles

R= .0821 (constant)

T= 273 K

1.00(6.20)=n(.0821)273

n= .277 moles CO2

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12y ago

According to Hyperphysics, at STP, carbon dioxide gas has a density of 0.001977g/mL

Divide the given mass by the known density.

6.5g / 0.001977g/mL = 3287.8mL*

*With correct significant figures, the answer should be rounded to 3300mL

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12y ago

(CO2)= 44 g/mol =molar weight

6.5g =mass

1L=1000ml

1mole=22.4L

n=m/mw 6.5g/44g/m=0.148mol CO2 = ( 0.148mole) (22.4L/1mole) (1000ml/1L) =3310ml

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6y ago

The volume of carbon dioxide is 3 288 mL.

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11y ago

6.23 moles x 44 grams/mole = 274.12 grams

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Q: What volume in milliters will 6.5g of CO2 occupy at stp?
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