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\nthe correct setup allows you to cancel terms to yield the mass of iron:\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}=? g fe\nwhat mass of fe forms? make sure to give your answer with the correct number of significant figures.\na total of 0.0257 g al forms g fe.\ndone\n0.1064\n1.435\n0.0532

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\nthe correct setup allows you to cancel terms to yield the mass of iron:\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}=? g fe\nwhat mass of fe forms? make sure to give your answer with the correct number of significant figures.\na total of 0.0257 g al forms g fe.\ndone\n0.1064\n1.435\n0.0532

Answer

Explanation:

Step1: Calculate moles of Al

$n_{Al}=\frac{m_{Al}}{M_{Al}}=\frac{0.0257\ g}{26.98\ g/mol}$

Step2: Use mole - ratio to find moles of Fe

From the balanced equation $Fe_2O_3 + 2Al\rightarrow Al_2O_3+2Fe$, the mole - ratio of $Al$ to $Fe$ is 2:2 or 1:1. So $n_{Fe}=n_{Al}\times\frac{2\ mol\ Fe}{2\ mol\ Al}$

Step3: Calculate mass of Fe

$m_{Fe}=n_{Fe}\times M_{Fe}=n_{Al}\times\frac{2\ mol\ Fe}{2\ mol\ Al}\times55.85\ g/mol=\frac{0.0257\ g}{26.98\ g/mol}\times\frac{2\ mol\ Fe}{2\ mol\ Al}\times55.85\ g/mol$ $m_{Fe}=\frac{0.0257\times55.85}{26.98}\ g\approx0.0532\ g$

Answer:

0.0532 g