Question

Quantos mols de compostos estão presentes em:
(a) 12,6 g de CdTe
(b) 27,0 g de C4H10
(c) 2,7 mols de C3H8O2N

Answer 1

Resposta:

a) 12,6 g de CdTe

Cd = 112,4  g/mol

Te = 127,6 g/mol

CdTe = 112,4 + 127,6 ⇒ 240 g/mol

1 mol → 240 g

x mol → 12,6 g

240x = 12,6

x= $\frac{12,6}{240}$

x=0,0525

---------//-----------

b) 27,0 g de C₄H₁₀

C = 12,0 g/mol

H = 1,0 g/mol

C₄H₁₀ = (12*4)+(1*4) ⇒ 52,0 g/mol

1 mol → 52,0 g

x mol → 27,0 g

52x = 27

x = $\frac{27}{52}$

x = 0,5192

----------//--------------

c) 2,7 mols de C₃H₈O₂N

C = 12,0 g/mol

H = 1,0 g/mol

O = 16,0 g/mol

N = 14,0 g/mol

C₃H₈O₂N = (12*3)+(1*8)+(16*2)+14 ⇒ 90 g/mol

1 mol → 90 g

2,7 mol → x g

x = 243 g

espero ter ajudado mano

1 mol é o mesmo que 6,02*10²³

então se quiser transformar só colocar na regrinha de 3 vlw