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

1 Answer
Jun 16, 2016

Answer:

#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 "#

#-----------------#

# 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#

#--------------------#

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#