What volume of H2 is formed at STP when 6.0 g of Al is treated with excess NaOH? 2NaOH + 2Al + 6H2O → 2NaAl(OH)4 + 3H2(g)

7.5

Explanation:

2NaOH+2Al+6H2O→2NaAl(OH)4+3H22NaOH+2Al+6H2O→2NaAl(OH)4+3H2

Since the sodium hydroxide is in excess, the aluminum is the limiting reagent. The moles of hydrogen produced therefore depend on the moles of aluminum fed into the reaction. Calculate the moles of aluminum by dividing the mass of aluminum by the molar mass of aluminum, 27.0 grams per mole.

6.0gAl27.0gmol=0.222molAl6.0gAl27.0gmol=0.222molAl

Looking at the stoichiometry, there are three moles of hydrogen gas produced for every two moles of aluminum. The moles of hydrogen produced are

0.222molAl×3molH22molAl=0.333molH20.222molAl×3molH22molAl=0.333molH2

Assuming that the hydrogen gas in an ideal gas, we can use the ideal gas law PV = nRT to find its volume. In this equation

∗P=pressure,atm∗V=volume,L∗n=amountofgas,mol∗R=gasconstant,0.0821LatmKmol∗T=temperature,Kelvin∗P=pressure,atm∗V=volume,L∗n=amountofgas,mol∗R=gasconstant,0.0821LatmKmol∗T=temperature,Kelvin

The standard temperature and pressure (STP) conditions are exactly one atmosphere and zero degrees Celsius. Convert the degrees Celsius to Kelvin by adding 273, so the temperature is 273 K. Now rearrange the equation to solve for the volume.

V=nRTP=0.333mol×0.0821LatmKmol×273K1atm=7.47LV=nRTP=0.333mol×0.0821LatmKmol×273K1atm=7.47L

Rounded to two significant figures, the volume of hydrogen gas produced is 7.5 liters.

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