consider again the thermite reaction. if 0.0257 g al react completely, what mass of fe forms? use the…

consider again the thermite reaction. if 0.0257 g al react completely, what mass of fe forms? use the periodic table to find molar masses.\nfe₂o₃ + 2al → al₂o₃ + 2fe\nselect the correct setup for this calculation.\n0.0257 g al×\\frac{26.98 g al}{1 mol al}×\\frac{2 mol al}{2 mol fe}×\\frac{55.85 g fe}{1 mol fe}\n0.0257 g al×\\frac{1 mol al}{26.98 g al}×\\frac{2 mol al}{2 mol fe}×\\frac{1 mol fe}{55.85 g fe}\n0.0257 g al×\\frac{1 mol al}{26.98 g al}×\\frac{2 mol fe}{2 mol al}×\\frac{55.85 g fe}{1 mol fe}

consider again the thermite reaction. if 0.0257 g al react completely, what mass of fe forms? use the periodic table to find molar masses.\nfe₂o₃ + 2al → al₂o₃ + 2fe\nselect the correct setup for this calculation.\n0.0257 g al×\\frac{26.98 g al}{1 mol al}×\\frac{2 mol al}{2 mol fe}×\\frac{55.85 g fe}{1 mol fe}\n0.0257 g al×\\frac{1 mol al}{26.98 g al}×\\frac{2 mol al}{2 mol fe}×\\frac{1 mol fe}{55.85 g fe}\n0.0257 g al×\\frac{1 mol al}{26.98 g al}×\\frac{2 mol fe}{2 mol al}×\\frac{55.85 g fe}{1 mol fe}

Answer

Explanation:

Step1: Convert mass of Al to moles

Use the molar - mass of Al ($M_{Al}=26.98\ g/mol$). The number of moles of Al ($n_{Al}$) is calculated by the formula $n=\frac{m}{M}$. So, $n_{Al}=\frac{0.0257\ g}{26.98\ g/mol}$, which is equivalent to $0.0257\ g\ Al\times\frac{1\ mol\ Al}{26.98\ g\ Al}$.

Step2: Use mole - ratio from the balanced equation

The balanced equation is $Fe_2O_3 + 2Al\rightarrow Al_2O_3+2Fe$. The mole - ratio of Al to Fe is $2:2 = 1:1$. So, the number of moles of Fe formed is the same as the number of moles of Al reacted considering the ratio $\frac{2\ mol\ Fe}{2\ mol\ Al}$.

Step3: Convert moles of Fe to mass

Use the molar - mass of Fe ($M_{Fe}=55.85\ g/mol$). The mass of Fe ($m_{Fe}$) is calculated by $m = n\times M$. So, $m_{Fe}=n_{Fe}\times55.85\ g/mol$, which is equivalent to multiplying the moles of Fe by $\frac{55.85\ g\ Fe}{1\ mol\ Fe}$. The correct setup is $0.0257\ g\ Al\times\frac{1\ mol\ Al}{26.98\ g\ Al}\times\frac{2\ mol\ Fe}{2\ mol\ Al}\times\frac{55.85\ g\ Fe}{1\ mol\ Fe}$.

Answer:

$0.0257\ g\ Al\times\frac{1\ mol\ Al}{26.98\ g\ Al}\times\frac{2\ mol\ Fe}{2\ mol\ Al}\times\frac{55.85\ g\ Fe}{1\ mol\ Fe}$