Cotton wool weighing 0.35 kg containing 92% cellulose was subjected to acid hydrolysis and received 0.116

Cotton wool weighing 0.35 kg containing 92% cellulose was subjected to acid hydrolysis and received 0.116 kg of glucose. Calculate the practical yield (%) of glucose.

Given:

m ((C6H10O5) n) = 0.35 kg = 350 g
W ((C6H10O5) n) = 92%
m (C6H12O6) = 0.116 kg = 116 g
η (C6H12O6) -?

DECISION:
(C6H10O5) n -> (H2O, H +) nC6H12O6

η = (m practice: m theory) * 100%
1) m (C6H12O6) = m ((C6H10O5) n) * w = 350 * 0.92 = 322 g
2) n (C6H12O6) = m (C6H12O6): M (C6H12O6) = 322: (6 * 12 + 1 * 10 + 16 * 5) = 322: 162 ~ 2 mol
3) n ((C6H10O5) n): n (C6H12O6) = 1: 1 => n (C6H12O6) = 2 mol
4) m theor. ch = n ch * M ch = 2 * (6 * 12 + 1 * 12 + 16 * 6) = 180 * 2 = 360 g
5) η (C6H12O6) = (m practice: m theory) * 100% = (116: 360) * 100% ~ 32%
Answer: 32%.