Cu =63.5
2 times N =28.02
6 times O =96
96=63.5=28.02=127.07
122/127.07=.96 molecules
Answer:
Mass = 357.7 g
Explanation:
Given data:
Mass of Fe = 250 g
Mass of oxygen = 120 g
Mass of iron(III) oxide produced = ?
Solution:
Chemical equation:
4Fe + 3O₂ → 2Fe₂O₃
Number of moles of Fe:
Number of moles = mass/molar mass
Number of moles = 250 g/ 55.8 g/mol
Number of moles = 4.48 mol
Number of moles of O₂ :
Number of moles = mass/molar mass
Number of moles = 120 g/ 32 g/mol
Number of moles = 3.75 mol
Now we will compare the moles of reactants with product.
Fe : Fe₂O₃
4 : 2
4.48 : 2/4×4.48 = 2.24
O₂ : Fe₂O₃
3 : 2
3.75 : 2/3×3.75= 2.5
Less number of moles of Fe₂O₃ are produced by Fe thus it will act as limiting reactant.
Mass of Fe₂O₃:
Mass = number of moles × molar mass
Mass = 2.24 mol × 159.69 g/mol
Mass = 357.7 g
Reaction won't occur
Because the activity of Cu is less than Al
