Question

Quantos mols de átomos estão presentes em uma mistura contendo 7,6 gramas de KOH; 134,1 gramas de NaI e 56 gramas de CaO?

Answer 1

K = 39g/mol
Na = 23g/mol
Ca = 40g/mol
H = 1g/mol
O = 16g/mol
I = 127g/mol

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KOH = 56g/mol
NaI = 150g/mol
CaO = 56g/mol

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6,02×10²³ → 56g KOH
x → 7,6g KOH

♦ x = (7,6 × 6,02×10²³)÷56
x = 0,817×10²³

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6,02×10²³ → 150g NaI
x → 134,1g NaI

♦x = (134,1 × 6,02×10²³)÷150
x = 5,38×10²³

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6,02×10²³ → 56g CaO
x → 56g CaO

♦x = (56 × 6,02×10²³)÷56
x = 6,02×10²³