By hydrolysis of maltose, glucose was obtained, weighing 13.5 g more than the mass of the initial disaccharide.

| January 9, 2021 | education

By hydrolysis of maltose, glucose was obtained, weighing 13.5 g more than the mass of the initial disaccharide. What volume of gas was released when a sufficient amount of metallic Na interacted with butane acid, obtained from glucose.

Given: Δm = 13.5 g
Find: V (gas) -?
Decision:
1) Write the reaction equations: C12H22O11 + H2O => 2C6H12O6; C6H12O6 => C3H7COOH + 2CO2 ↑ + 2H2 ↑; 2C3H7COOH + 2Na => 2C3H7COONa + H2 ↑; 2) Find the mass of H2O: Since only 2 substances take part in the hydrolysis of maltose, and it turns out 1, the difference in masses will be nothing more than the mass of H2O (according to the law of conservation of mass of substances). Then m (H2O) = Δm = 13.5 g; 3) Find the amount of substance H2O: n (H2O) = m (H2O) / Mr (H2O) = 13.5 / 18 = 0.75 mol; 4) Find the amount of substance H2 (taking into account the reaction equations): n (H2) = n (2C3H7COOH) / 2 = n (C6H12O6) / 2 = n (H2O) = 0.75 mol; 5) Find the volume of gas H2: V (H2) = n (H2) * Vm = 0.75 * 22.4 = 16.8 liters.
Answer: The volume of H2 gas evolved is 16.8 liters.



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