How Much Magnesium is Needed for 40 mL of Hydrogen Gas Production?

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SUMMARY

The discussion focuses on calculating the grams of magnesium (Mg) required to produce 40 mL of hydrogen gas (H2) when reacted with hydrochloric acid (HCl) at 22 degrees Celsius and 1.02 bar. Using the ideal gas law (PV=nRT), the calculation yields approximately 0.040 grams of Mg needed. Additionally, with a magnesium ribbon density of 0.836 mg/mm, approximately 47.85 mm of the ribbon is required for the reaction. The participants emphasize the importance of unit consistency and pressure conversion in gas law calculations.

PREREQUISITES
  • Understanding of the Ideal Gas Law (PV=nRT)
  • Basic knowledge of stoichiometry in chemical reactions
  • Familiarity with unit conversions, particularly pressure (atm to bar)
  • Knowledge of molar mass calculations (e.g., molar mass of Mg)
NEXT STEPS
  • Research the Ideal Gas Law applications in various chemical reactions
  • Learn about stoichiometric calculations for reactants and products
  • Explore pressure conversion techniques between different units
  • Study the properties and applications of magnesium in chemical reactions
USEFUL FOR

Chemistry students, educators, and professionals involved in chemical reaction calculations and gas law applications will benefit from this discussion.

kirsten_2009
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Homework Statement



Determine how many grams of Mg must react with HCl in order to produce 40 mL of hydrogen gas at 22 degrees Celsius and 1.02 bar. Also, If the Mg ribbon weighs 0.836 mg/mm, how many mm of the ribbon will you require?

Homework Equations



PV=nRT

Mg(s) + 2HCl(aq) --> MgCl2(aq) + H2(g)

The Attempt at a Solution



n=PV/RT
n= (1.00 atm)(0.04 L) / (0.0821 L*atm/mol*K)(295.15 K)
n= 0.04/24.232 = 0.0016507098 mol * 24.3050 g/mol = 0.040 g Mg

40 mg x 1mm / 0.836mg = 47.85mm

Correct? Thanks for the help!
 
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Result looks OK to me, calculations are rather confusing.

kirsten_2009 said:
n= (1.00 atm)(0.04 L) / (0.0821 L*atm/mol*K)(295.15 K)

Let's say it is OK, although 1 atm is not exactly equal to 1.02 bar.

n= 0.04/24.232 = 0.0016507098 mol * 24.3050 g/mol

And that's where you have lost me. What is 0.04 and where did you get it from? What is 24.232 and where did you get it from? They have no units, so I would have to guess what they are, plus, they don't follow from the previous line.
 
Hello,

Yes, I see what you are saying. I will add the units and be more specific with my pressure conversion.

the 0.04 comes from 0.04L x 1.00 atm and the 24.232 comes from (0.0821 L*atm/mol*K)*(295.15 K). Thanks for your help! :)
 

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