Question

Fórmula mínima de 63.1% O 36%N
Y 21.6%Mg, 12.4%C, 57.0% O
Porfa ayuda!!!

Answer 1

Respuesta:

A) N2O3

B) MgCO3

Explicación:

Fórmula mínima de

a. 63.1% O 36%N y

b. 21.6%Mg, 12.4%C, 57.0% O

a. calcular moles de cada elemento (63.1% O 36% N)

n O = 63.1 g / 16 g/mol = 3.94 mol

n N = 36 g / 14 g/mol = 2.57 mol

dividir entre el menor de los resultados

N: 2.57 mol / 2.57 mol = 1 x 2 = 2

O: 3.94 mol / 2.57 mol = 1.5 x 2 = 3

multiplicar los resultados por 2 uno tiene decimal

FM: N2O3

b. calcular moles de cada elemento (21.6%Mg, 12.4%C, 57.0% O)

n Mg = 21.6 g / 24 g/mol = 0.9 mol

n C = 12.4 g / 12 g/mol = 1.03 mol

n O: 57.0 g / 16 g/mol = 3.56 mol

dividir cada resultado entre el menor

Mg: 0.9 mol / 0.9 mol = 1

C: 1.03 mol / 0.9 mol = 1

O: 3.56 mol / 0.9 mol = 4

FE: MgCO4

Nota; al sumar los % falta para llegar al 100%

``````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````

Calcular moles de cada elemento (28.8% Mg, 14.2% C, 56.93% O)

n Mg = 28.8 g / 24 g/mol = 1.2 mol

n C = 14.2g / 12 g/mol = 1.2 mol

n O: 56.93 g / 16 g/mol = 3.56 mol

dividir cada resultado entre el menor

Mg: 1.2 mol / 1.2mol = 1

  C: 1.2 mol / 1.2 mol = 1

  O: 3.56 mol / 1.2 mol = 3

FE: MgCO3