What is the Molar Ratio for Dissolving Aluminum with Sulfuric Acid?

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SUMMARY

The molar ratio for dissolving aluminum with sulfuric acid is established by the balanced chemical equation: 2 Al(s) + 3 H2SO4 → Al2(SO4)3(aq) + 3 H2. This indicates that 3 moles of sulfuric acid are required for every 2 moles of aluminum. To dissolve a 15.2 g aluminum block, one must calculate the minimum mass of H2SO4 needed using the molar masses of aluminum (26.98 g/mol) and sulfuric acid (98.09 g/mol). The correct calculation involves determining the moles of aluminum and applying the molar ratio to find the corresponding mass of sulfuric acid.

PREREQUISITES
  • Understanding of stoichiometry
  • Familiarity with molar mass calculations
  • Knowledge of balanced chemical equations
  • Basic algebra for mass and mole conversions
NEXT STEPS
  • Calculate the moles of aluminum in 15.2 g
  • Determine the mass of sulfuric acid required using the molar ratio
  • Review stoichiometric calculations in chemical reactions
  • Explore practical applications of sulfuric acid in metal dissolution
USEFUL FOR

Chemistry students, educators, and anyone involved in chemical engineering or laboratory work focusing on acid-base reactions and metal dissolution processes.

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Homework Statement



sulfuric acid dissolves aluminum metal according to
2 Al(s) + 3 H2SO4 ---- Al2(SO4)3 (aq) + 3 H2

find minimum mass of H2SO4 in g if you wanted to dissolved aluminum block with mass of 15.2 g

Homework Equations



see above

The Attempt at a Solution


15.2 g Al x 1/26.98g x I am confused about molar ratios what do i use? x 98.09 g H2So4

could someone help me with molar ratios i don't understand what to multiply by
 
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Look at the reaction equation - how many moles of acid needed for 2 moles of aluminum?
 

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