What is the mass of silver released during the oxidation of 230 g of glucose containing 11% impurities

What is the mass of silver released during the oxidation of 230 g of glucose containing 11% impurities if an ammonia solution of silver oxide (Ag2o) serves as an oxidizing agent.

1. We make a chemical equation, not forgetting to put down the coefficients:
C6H12O6 + Ag2O = CH2OH – (CHOH) 4 – COOH + 2Ag.
2. Find the mass of C6H12O6:
m (C6H12O6) = 180 g / mol * 1 mol = 180g.
3. Find the mass Ag:
m (Ag) = 108 g / mol * 2 mol = 216 g.
4. Find the mass of impurities:
m (impurities) = 11% * 230g / 100% = 25.3g.
5. Find the mass of glucose:
m (glucose) = 230g – 25.3g = 204.7g.
6.Using the equation, compose the proportion and find the mass of silver:
204.7g / 180g = Xg / 216g.
m (Ag) = 204.7g * 216g / 180g = 245.64g.
Answer: m (Ag) = 245.64g.