Question

Need help for chemistry

Answer 1

Answer:

• a) Initial mass of irong oxide: 6,310g

• b) Initial mass of Aluminum: 2,170g

Explanation:

It is necessary to use the percent yield of the experiment. Since it is not provided, I will use a hypothetical percent yield of 90%.

1. Write the chemical equation:

      $Fe_2O_3+2Al\longrightarrow 2Fe+Al_2O_3$

2. Write the mole ratios:

      $\dfrac{1molFe_2O_3}{2molFe}\\ \\ \\ \dfrac{2molAl}{2molFe}$

3. Covert 5.00 kg of soid iron to number of moles (n)

• n = mass in grams / atomic mass

• atomic mass of Fe = 55.845g/mol

• (5.00kg × 1,000g/kg) ÷ (55.845 g/mol) = 89.5335mol Fe

4. Use the mole ratios to find the number of moles of each reactant

      $89.5335molFe\times \dfrac{1molFe_2O_3}{2molFe}=44.76675molFe_2O_3$

      $89.5335molFe\times \dfrac{2molAl}{2molFe}=89.5335molAl$

5.  Use the molar masses to convert the number of moles into mass:

• mass = number of moles × molar mass

Fe₂O₃:

• molar mass: 159.69 g/mol

• mass = 44.76675mol × 156.69g/mol = 7,014.50g

Al:

• atomic mass: 26.982g/mol

• mass = 89.5335mol × 26.982g/mol = 2,415.79g

6. Multiply by the percent yield: 90% (assumed)

Fe₂O₃:

• 7,014.50g × 90% = 6,313.05g = 6,310g (three significant figures)

Al:

• 2,415.79g × 90% = 2,174.21g = 2,170g (three significant figures)