How to Solve Stoichiometry Problems Involving Mass and Balanced Equations?

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SUMMARY

The discussion focuses on solving stoichiometry problems involving mass and balanced chemical equations, specifically using the reaction 2Al(NO3)3 + 3Mg ==> 3Mg(NO3)2 + 2Al. The user successfully converted 54.1 g of aluminum nitrate into approximately 0.2539 moles and learned to apply the mole ratio of 2:3 between aluminum nitrate and magnesium nitrate. The final solution involved using dimensional analysis to calculate the mass of magnesium nitrate produced from the moles of aluminum nitrate.

PREREQUISITES
  • Understanding of stoichiometry and mole conversions
  • Familiarity with balanced chemical equations
  • Knowledge of dimensional analysis techniques
  • Ability to perform mole-to-mass conversions
NEXT STEPS
  • Study the concept of mole ratios in chemical reactions
  • Learn about dimensional analysis in chemistry
  • Practice converting grams to moles and vice versa
  • Explore more complex stoichiometry problems involving multiple reactants and products
USEFUL FOR

Chemistry students, educators, and anyone looking to improve their skills in stoichiometry and chemical reaction calculations.

GLprincess02
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Homework Statement



Given the balanced equation 2Al(NO3)3 + 3Mg ==> 3Mg(NO3)2 + 2Al , if 54.1 g of aluminum nitrate is reacted, what mass of magnesium nitrate is produced?

2. The attempt at a solution
I've been given an entire packet dedicated to stoichiometry, but nothing in here covers this problem. I am unsure of how to even begin solving it. Maybe if I had help with the first few steps I could solve the rest.
 
Last edited:
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a good way to go about this problem is to convert the mass of aluminum nitrate into moles of aluminum nitrate, then you can figure out how many moles of magnesium (do you mean aluminum or magnesium nitrate) were produced.
 
Ok so I converted the grams of aluminum nitrate into moles and got approx. 0.2539 moles. I'm still confused about what to do next, though.
 
Last edited:
GLprincess02 said:
Ok so I converted the grams of aluminum nitrate into moles and got approx. 0.2539 moles. I'm still confused about what to do next, though.

Now you need to use the mole fraction, which you get from the balanced reaction.
 
What is the mole fraction? Like, 2 moles of aluminum nitrate to 3 moles of magnesium nitrate?
 
GLprincess02 said:
What is the mole fraction? Like, 2 moles of aluminum nitrate to 3 moles of magnesium nitrate?

Yes.
(2mol Al nitrate/3mol Mg nitrate)
 
Ok, so how do I use this ratio in relation to the 0.2539 moles of aluminum nitrate? I guess I still don't really get all of the steps.
 
GLprincess02 said:
Ok, so how do I use this ratio in relation to the 0.2539 moles of aluminum nitrate? I guess I still don't really get all of the steps.

It is a dimensional analysis setup

mol Al nitrate(mole fraction)(convert mol Mg nitrate to g)=soln
 
So I take 0.2539 and multiply it by 2/3 first, right?
 
  • #10
Never mind, I got the right answer. Thanks for the help. :)
 

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