Respuesta :

ardni313

The mass of iron (III) oxide (Fe2O3) : 85.12 g

Further explanation

Given

3.20x10²³ formula units

Required

The mass

Solution

1 mole = 6.02.10²³ particles  

Can be formulated :

N = n x No

N = number of particles

n = mol

No = 6.02.10²³ = Avogadro's number

mol of Fe₂O₃ :

[tex]\tt n=\dfrac{3.2.10^{23}}{6.02.10^{23}}=0.532[/tex]

mass of Fe₂O₃ (MW=160 g/mol)

[tex]\tt mass=mol\times MW=0.532\times 160=85.12~g[/tex]

We have 84.84g of  Fe[tex]_2[/tex]O[tex]_3[/tex] in 3.20*10^23 particles.

Data;

  • number of particles = 3.20*10^23
  • molar mass = 159.69g/mol
  • Avogadro's number = 6.023*10^23 particles.

To solve this problem, we need to relate Avogadro number to the molar mass.

From molar mass - Avogadro's number relationship;

[tex]159.69g = 6.023*10^2^3\\xg = 3.20*10^2^3\\x = \frac{159.69*3.20*10^2^3}{6.023*10^2^3} \\x = 84.84g[/tex]

From the above calculations, 84.84g of Fe[tex]_2[/tex]O[tex]_3[/tex] is present in 3.20*10^23 particles of  Fe[tex]_2[/tex]O[tex]_3[/tex].

Learn more on Avogadro's number here;

https://brainly.com/question/20091306

Otras preguntas