105)
a)
mass of Fe = 15 g
molar mass of Fe = 55.85 g/mol
mol of Fe = (mass)/(molar mass)
= 15/55.85
= 0.268577 mol
According to balanced equation
mol of Fe2O3 required = (1/2)* moles of Fe
= (1/2)*0.268577
= 0.134288 mol
mass of Fe2O3 = number of mol * molar mass
= 0.134288*215.55
= 28.9 g
Answer: 28.9 g
b)
According to balanced equation
mol of Al required = moles of Fe
= 0.268577 mol
mass of Al = number of mol * molar mass
= 0.268577*26.98
= 7.25 g
Answer: 7.25 g
c)
According to balanced equation
mol of Al2O3 formed = (1/2)* moles of Fe
= (1/2)*0.268577
= 0.134288 mol
mass of Al2O3 = number of mol * molar mass
= 0.134288*128.94
= 17.3 g
Answer: 17.3 g
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Reaction Stoichiometry os Over the years, the thermite reaction has been used for weld- ing railroad...
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