m tech. (Ca) = 8 g
ω approx. = 30%
n (CaO) -?
1) 2Ca + O2 => 2CaO;
2) ω (Ca) = 100% – ω approx. = 100% – 30% = 70%;
3) m clean. (Ca) = ω (Ca) * m tech. (Ca) / 100% = 70% * 8/100% = 5.6 g;
4) n (Ca) = m (Ca) / M (Ca) = 5.6 / 40 = 0.14 mol;
5) n (CaO) = n (Ca) = 0.14 mol.
Answer: The amount of CaO substance is 0.14 mol.
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