By hydrolysis of 250 kg of sawdust, the cellulose content of which is 45%, 62 kg

By hydrolysis of 250 kg of sawdust, the cellulose content of which is 45%, 62 kg of glucose were obtained. Determine the mass fraction of the yield of glucose from the theoretically possible.

Given:
m (sawdust) = 250 kg = 250,000 g
ω ((C6H10O5) n) = 45%
m pract. (C6H12O6) = 62 kg = 62000 g

To find:
ω out. -?

Decision:
1) (C6H10O5) n + nH2O => nC6H12O6;
2) m ((C6H10O5) n) = ω ((C6H10O5) n) * m (sawdust) / 100% = 45% * 250000/100% = 112500 g;
3) n ((C6H10O5) n) = m ((C6H10O5) n) / M ((C6H10O5) n) = 112500 / 162n = 694.4n mol;
4) n (C6H12O6) = n ((C6H10O5) n) * n = 694.4n * n = 694.4 mol;
5) m theor. (C6H12O6) = n (C6H12O6) * M (C6H12O6) = 694.4 * 180 = 124992 g;
6) ω out. = m practical (C6H12O6) * 100% / m theor. (C6H12O6) = 62000 * 100% / 124992 = 49.6%.

Answer: The mass fraction of the output of C6H12O6 is 49.6%.