# If 2.0 grams of hydrogen gas was occupied at STP, determine the volume in cubic meters.

To solve this problem, the Ideal gas equation is important. Remember that the ideal gas law states that:

`PV = nRT`

to solve for the volume of Hydrogen gas, we can arrange the equation into:

`V = (nRT)/(P)`

Given:

Mass `H_2` = 2.0 grams

P = 0.986 atm (at STP)

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To solve this problem, the Ideal gas equation is important. Remember that the ideal gas law states that:

`PV = nRT`

to solve for the volume of Hydrogen gas, we can arrange the equation into:

`V = (nRT)/(P)`

Given:

Mass `H_2` = 2.0 grams

P = 0.986 atm (at STP)

T = 273.15 K (at STP)

R = gas constant = 0.08206 `(atm-L)/(mol-K)`

First, we get the number of moles of Hydrogen gas.

`(2.0 grams H_2)* (1 mol e H_2)/(2.016 grams H_2)`

= 0.9921 moles `H_2`

Now we are ready to get the value of the volume.

V = (nRT)/(P)

V = (0.9921*0.08206*273.15)/(0.986atm)

V = 22.6 Liters of `H_2`

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The unit of volume can be expressed in many ways. I can also be equal to:

22600 mL; 22600`cm^3` ; 0.0226 cubic meters

note: There are many interpretations of STP, some would say that at STP, the condition of a gas is at 1atm (not 0.986at,) and 273.15 K.