Respuesta :
Mass of iron(III)oxide, Fe2O3 sample = 1.143*10^2 g
Molar mass of Fe2O3 = 159.69 g/mole
i.e. 1 mole of Fe2O3 weighs 159.69 g
Based on stoichiometry:
1 mole of Fe2O3 has 2 moles of Fe
Atomic mass of Fe = 55.85 g/mole
Therefore we can write:
159.69 g of Fe2O3 is composed of 2*55.85 g of Fe
Thus, amount of Fe present in 1.143*10^2 g sample of Fe2O3 is
= 1.143*10^2 * 2* 55.85/159.69
= 7.995*10^-1 g
79.86 grams of iron are present in 1.143×10²g sample of iron(III) oxide Fe₂O₃.
How we calculate mass from?
Mass of any substance can be calculated by using their moles as:
n = W/M, where
W = given mass
M = molar mass
According to the stoichiometry:
1 mole of Fe₂O₃ = contain 2 mole of Fe
In the question given that mass of Fe₂O₃ = 1.143×10²g
Mole of Fe₂O₃ = 1.144×10²g / 159.69 g/mole = 0.715mole
0.715 mole of Fe₂O₃ = contain 2×0.715 mole of Fe
So, mole of Fe = 1.43 mole
Now required mass of Fe will be calculated by using the above formula of mole as:
W = n × M
W = 1.43mole × 55.85g/mole = 79.86g
Hence, 79.86g of iron are present.
To know more about moles, visit the below link:
https://brainly.com/question/14464650
