CHEMISTRY HELP PLS

Question

How many liters of hydrogen gas at STP can be generated from the reaction of excess hydrogen chloride with 79.5g of aluminum?

Answer 1

Hi, Molly,

'Happy to help you here! First, we need to write a balanced equation for the reaction of hydrogen chloride (HCl, also known as hydrochloric acid) with aluminum (simply Al):

2Al + 6HCl -> 2AlCl₃ + 3H₂

First, as you can see/check for yourself, we have the same number of Al, H, and Cl atoms on each side of the arrow/equation, so this is now balanced!

Next, we need to convert 79.5 grams of aluminum into units we can actually work with - that would be moles! In the case of aluminum (based on its atomic weight), 1 mole of Al weighs 27.0 grams. We can now solve the balanced equation to determine the number of moles of hydrogen produced:

79.5 g Al/27.0 g per mole = 2.94 moles Al

Since the equation states that 2Al + 6HCl -> 2AlCl₃ + 3H₂, and the HCl is in excess, we can solve for hydrogen as follows:

From this equation, we see that 2Al will produce 3H₂, therefore

2.94 moles Al will produce 3/2 (2.94) or 4.41 moles H₂.

Finally, we need to invoke the Ideal Gas Equation to convert moles of a gas into the actual volume of gas at standard temperature (273.15K) and pressure (1 atm):

PV = nRT, or

V = nRT/P, where n is the number of moles of gas, R is the molar gas constant, and T is the temperature (in degrees Kelvin, or K)

For our problem,

V = (4.41 moles H²)(0.0821 L^.atm/K^.mol)(273.15 K)/1 atm, or

V = 4.41 x 0.0821 x 273.15, or

V = 99.0 liters H₂ at STP

'Hope that helps,

Chris