How many liters are in 16 grams of H_2 at STP?

1 Answer
Jun 16, 2016

V~=1.8xx10^2\ L

Explanation:

Use the ideal gas equation

P*V = n * R*T

Where:

P -> " is the pressure expressed in " atm

V -> " is the volume occupied by the gas expressed in "L

n-> " is the number of moles of the gas"

R ->" is the universal gas constant" =0.0821 \ L* atm*mol^-1 *K^-1

T -> " is the kelvin temperature "

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S.T.P "conditions" => T = 273 K and P= 1.00 \ atm

Rearrange the formula and solve for V.

V = (n * R*T)/P

Find the number of moles of the hydrogen gas present in the 16 grams.

n_(H_2)= 16 \ g \ H_2xx(1 \ mol. H_2)/(2.016\ g \ H_2)

n_(H_2)~=7.9 \ mol.

V=(7.9 \ mol. xx 0.0821 \ L* atm*mol^-1 *K^-1xx 273 \ K)/ (1.00 \ atm)

V=(7.9 \ cancel(mol.) xx 0.0821 \ L* cancel(atm)*cancel(mol^-1)*cancel(K^-1)xx 273 \ cancel(K))/ (1.00 \ cancel(atm))

V~=1.8xx10^2\ L

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A quick approach

At S.T.P you can use the following formula:

n_(H_2)= V/V_M

n_(H_2) " is the number of moles of the gas."

V " is Volume of the gas under S.T.P conditions"

V_M" is the molar volume i.e the volume occupied by 1 mole of "
"any gas under S.T.P conditions, equal to 22.4 L/mol."

n_(H_2)= 16 \ g \ H_2xx(1 \ mol. H_2)/(2.016\ g \ H_2)

n_(H_2)~=7.9 \ mol.

V = n_(H_2)xxV_M

V = 7.9 \ mol xx 22.4 L* mol^-1

V~=1.8xx10^2\ L