# Suppose you ran this same reaction (using the balloon setup as seen

Suppose you ran this same reaction (using the balloon setup as seen
in the video) on your own with two different flasks. In Flask A, you reacted 5.10 g Mg with 0.447 mol HCl. In Flask B, you reacted 24.21 g Mg with 0.998 mol HCl. Which ballon will inflate the most? Explain (in detail) why, showing all work.
Mg + 2HCl = MgCl2 + H2
This is the anwser to the above question:
In this reaction the ratio of reactants is 1:2.
In Flask A – 5.10 g/ (24.305 g/mol) = 0.20983 mol Mg
In Flask B – 24.21g/(24.305 g/mol) = .99607 mol Mg
Flask A – .20983 mol Mg/.447 mol HCl < ½
Mg is limiting reactant
Flask B – .99607 mol Mg/.998 mol HCl > ½
HCl is limiting reactant
Balloon A – ratio of Mg to H2 is 1:1
.20983 mols H2 generated
Balloon B – ratio of HCl to H2 is 2:1
.998 mols HCl/2 = .499 mols H2 generated
Balloon B inflates the m
How do you find the mol ratio what is the 24.305 g/mol where would i get that info? i see the math but dont know where the formula numbers came from

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