Question

Какой максимальный объём водорода(измеренного при н.у.)может быть выделен натрием из глицерина,взятого в количестве 1 моль????

Answer 1

             1                                                                                     x

2HOCH2-CH(OH)-CH2OH +6Na => 2NaOCH2-CH(ONa)-CH2(ONa) + 3H2

             2                                                                                      3
Решаем пропорцию 1/2=х/3 (числитель - моли вещества, знаменатель - коэффициент уравнения)
x=1.5
V=n*Vm= 1.5*22.4=33.6 моль водорода

Answer 2

Maximum Volume of Hydrogen Released from Glycerin by Sodium at STP

When sodium reacts with glycerin, it can produce hydrogen gas. The maximum volume of hydrogen gas that can be produced from 1 mole of glycerin at standard temperature and pressure (STP) can be calculated using the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between sodium and glycerin is as follows: ``` C3H8O3 + 3Na → 3CH3OH + 3/2H2 ``` From the balanced equation, it can be inferred that 1 mole of glycerin (C3H8O3) reacts with 3 moles of sodium (Na) to produce 3/2 moles of hydrogen gas (H2).

Therefore, the maximum volume of hydrogen gas released from 1 mole of glycerin can be calculated using the ideal gas law: ``` PV = nRT ``` Where: - P = pressure (at STP, P = 1 atm) - V = volume - n = number of moles - R = ideal gas constant (0.0821 L·atm/mol·K) - T = temperature (at STP, T = 273 K)

Using the ideal gas law, the maximum volume of hydrogen gas released from 1 mole of glycerin can be calculated as follows: ``` V = nRT/P = (3/2) * 0.0821 * 273 / 1 ≈ 33.4 L ``` Therefore, the maximum volume of hydrogen gas released from 1 mole of glycerin at STP is approximately 33.4 liters.