Calculating Mass from Chemical Reaction

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SUMMARY

The mass of alum that can be prepared from 1.340g of aluminum (Al) is calculated to be 23.55g. The reaction equation used is 2Al + 2KOH + 4H2SO4 + 22H2O → 3H2 + 2KAl(SO4)2·12H2O, where aluminum is identified as the limiting reagent. The moles of aluminum were determined using the formula n = m/M, resulting in 0.049662 mol of Al. The molar mass of alum is confirmed to be 474.3 g/mol, leading to the final mass calculation of alum produced.

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  • Knowledge of molar mass calculations
  • Familiarity with limiting reagents in chemical equations
  • Basic algebra for mole and mass conversions
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Homework Statement


what is the mass of alum that can be prepared from 1.340g of Al.

Homework Equations


2Al+2KOH+4H2SO4+22H2O ---> 3H2 + 2KAl(SO4)2 * 12H2O


The Attempt at a Solution


given Al is the limiting reagent. Find the moles of Al using n=m/M

I find Al moles to be 0.049662mol.

Now multiply this by the molar ratio of 2/2 (which is 1) to find the moles of alum produced.

now calculate for the mass, m = nM. The molar mass of alum is 474.3g/mol. I find the mass of alum to be 23.55g.

Am I right?
 
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