What mass of glucose can be obtained from starch weighing 81 grams if the mass fraction of the yield is 0.75?

Data: mC6H10O5 – weight of taken starch (mC6H1O5 = 81 g); ηC6H12O6 – mass fraction of the output (ηC6H12O6 = 0.75).
Const: MC6H12O6 – molar mass of glucose (MC6H12O6 = 180.16 g / mol); MC6H10O5 – molar mass of starch (MC6H10O5 = 162.141 g / mol).
1) The reaction level: (C6H10O5) n (starch) + nH2O (water) → (H2SO4, presence of sulfuric acid) nC6H12O6 (glucose).
2) Theoretical mass of glucose: mC6H12О6 / MC6H12О6 = νC6H12О6 = νC6H10О5 = mC6H10О5 / MC6H10О5, whence we express: mC6H12О6 = MC6H12О6 * mC6H10О5 / MC6H10О5 = 180.16 * 81/16 g.
3) The resulting mass of glucose: mpr = mC6H12O6 * ηC6H12O6 = 90 * 0.75 = 67.5 g.
Answer: From 81 g of starch, 67.5 g of glucose can be obtained.