Question

2 Calculează masa de oxigen care se gaseste in 15 moli de:
a) oxid de aluminiu (alumină), Al203
b) hidroxid de bariu (barita) Ba(OH)2
c) acid azotic (apă tare) HNO3
d) clorură de var, CaOCl2:​

Answer 1

Răspuns:

a) 720g oxigen

b) 480g oxigen

c) 720g oxigen

d) 240g oxigen

Explicație:

a)

M(Al₂O₃) = 2 × A(Al) + 3 × A(O) = 2 × 27 + 3 × 16 = 102 g/mol

1 mol Al₂O₃..........48g O

15 mol Al₂O₃..........x g O

x = 15 × 48 = 720g O

b)

M(Ba(OH)₂) = A(Ba) + 2 × A(O) + 2 × A(H) = 137 + 2 × 16 + 2 × 1 = 171 g/mol

1 mol Ba(OH)₂..........32g O

15 mol Ba(OH)₂..........y g O

y = 15 × 32 = 480g O

c)

M(HNO₃) = A(H) + A(N) + 3 × A(O) = 1 + 14 + 3 × 16 = 63 g/mol

1 mol HNO₃..........48g O

15 mol HNO₃..........z g O

z = 15 × 48 = 720g O

d)

M(CaOCl₂) = A(Ca) + A(O) + 2 × A(Cl) = 40 + 16 + 2 × 35,5 = 127 g/mol

1 mol CaOCl₂..........16g O

15 mol CaOCl₂..........w g O

w = 15 × 16 = 240g O

-Mihai