Respuesta :

SvetkaChem

Answer:

4.1646 kg

Explanation:

BaCl2

M(BaCl2)=137.33 + 2*35.45 = 208.23 g/mol

208.23 g/mol * 20 mol=4164.6 g ≈ 4.1646 kg

rosavo0414
BaCl2: 137 + 35.5 x 2 = 208 g/mol
20 mol x 208 g/mol = 4160 g BaCl2
1 g = 0.001 kg => 4160 g = 4.16 kg BaCl2