January 16, 2022, 05:33:29 AM
Forum Rules: Read This Before Posting

0 Members and 1 Guest are viewing this topic.

#### baseg_18

• New Member
• Posts: 8
• Mole Snacks: +1/-1
##### Question about a Single Displacement Reaction?
« on: June 15, 2006, 02:03:29 PM »
If 75.5 g of aluminum react with excess hydrochloric acid (hydrogen chloride), how much hydrogen gas will be produced in a single displacement reaction?

If it's not too much trouble can someone please explain the steps to solve this reaction problem, so that I can do these problems on my own.

Thanks,
Baseg_18

#### tamim83

• Retired Staff
• Full Member
• Posts: 395
• Mole Snacks: +67/-7
• Gender:
• Quantum Kitten
##### Re: Question about a Single Displacement Reaction?
« Reply #1 on: June 15, 2006, 03:37:16 PM »
Well the first thing you need is a chemical equation.  With problems like this that already does not give you one, you always want to start out with an equation.  Remember to balance this equation too so that you have proper molar relationships (I'll leave this up to you to figure out  ) After that, it is a fairly straight foward stoiciometry problem.  Convert the mass of Al to moles and use your balanced chemical equation to convert from moles of Al to moles of H2.  Then you can convert your moles of H2 to grams.

#### kayamusty

• Regular Member
• Posts: 16
• Mole Snacks: +1/-5
##### Re: Question about a Single Displacement Reaction?
« Reply #2 on: June 15, 2006, 05:04:14 PM »
Al + HCl -------- AlCl3  +  3/2 H2

use this equation

#### baseg_18

• New Member
• Posts: 8
• Mole Snacks: +1/-1
##### Re: Question about a Single Displacement Reaction?
« Reply #3 on: June 15, 2006, 05:06:09 PM »
Quantum Kitten or anybody,

Is this the correct way to do this problem?:

1st step-
Al + 2 HCl --> H2 + AlCl
(75.5 g)

2nd step-
75.5g Al * (1 mol Al)(1 mol H2) / (26.98g Al)(1 mol Al)

3rd step
75.5 g * (1 mol Al)(1 mol H2)(22.4 L H2) / (26.98g Al)(1 mol Al)(1 mol H2) = 62.6 L or 62.6 dm^3

#### wereworm73

• Chemist
• Full Member
• Posts: 179
• Mole Snacks: +21/-4
##### Re: Question about a Single Displacement Reaction?
« Reply #4 on: June 15, 2006, 05:25:49 PM »
You need to use this balanced equation:

Al + 3 HCl ---> AlCl3 + 3/2 H2
(Aluminum is trivalent so it bonds with three chloride ions to form AlCl3)

Then you figure out how many moles of Al are in 75.5 g and then just multiply that number by 3/2 (because by the balanced equation above, you get 3/2 moles of H2 for every mole of aluminum).  Then you'll have the number of moles of H2 gas.  If you need the amount of hydrogen expressed in liters, just multiply by 22.4 L/mol (assuming the pressure is 1 atm).
« Last Edit: June 15, 2006, 05:37:31 PM by wereworm73 »

#### baseg_18

• New Member
• Posts: 8
• Mole Snacks: +1/-1
##### Re: Question about a Single Displacement Reaction?
« Reply #5 on: June 15, 2006, 05:43:44 PM »
so would I multiply 62. 6 dm^3 * 1.5 (3/2) to get the answer 93.9 dm^3 ?

#### baseg_18

• New Member
• Posts: 8
• Mole Snacks: +1/-1
##### Re: Question about a Single Displacement Reaction?
« Reply #6 on: June 15, 2006, 05:55:36 PM »
I think I finally got it!

2.798 moles of Al * 1.5 = 4.197

4.197 * 22.4 = 94.0128

Thank you Wereworm73, Kayamusty, and Tamim83 for all your help.

#### tamim83

• Retired Staff
• Full Member
• Posts: 395
• Mole Snacks: +67/-7
• Gender:
• Quantum Kitten
##### Re: Question about a Single Displacement Reaction?
« Reply #7 on: June 16, 2006, 09:08:34 AM »
Quote
I think I finally got it!

2.798 moles of Al * 1.5 = 4.197

4.197 * 22.4 = 94.0128

Looks great!!

Quote
Thank you Wereworm73, Kayamusty, and Tamim83 for all your help.

No problem