What mass of silver is produced in the "silver mirror" reaction when an excess of glucose reacts with 50 g of a 3.4% silver nitrate solution?
- 1.08 g
- 2.16 g
- 0.54 g
- 4.32 g
📝 Detailed explanation
Mass AgNO₃ = 50 * 0.034 = 1.7 g.n(AgNO₃) = 1.7 / 170 = 0.01 mol.
In this reaction, 1 mole of AgNO₃ yields 1 mole of Ag.
Mass Ag = 0.01 * 108 = 1.08 g.