Question

Чему равен объем (н.у.) водорода, выделившегося в реакции между этанолом, хим. количеством 0,5 моль и натрием, хим.количество равно 0,009 моль?

Answer 1

2C2H5OH +2Na=2C2H5ONa + H2
По уравнению реакции количество этанола и натрия одинаково (2:2 или 1:1). По условию количество спирта больше, а натрия меньше. Решаем через меньшую величину.По уравнению реакции соотношение натрия и водорода 2:1. Значит количество водорода равно 0,09:2 = 0,045 моль. Объем водорода 0,045 * 22,4=

Answer 2

Calculating the Volume of Hydrogen Gas Produced

To calculate the volume of hydrogen gas produced in the reaction between ethanol and sodium, we can use the balanced chemical equation and the stoichiometry of the reaction.

The balanced chemical equation for the reaction is: C2H5OH + 2Na → 2C2H5ONa + H2

Given: - Chemical quantity of ethanol (C2H5OH) = 0.5 moles - Chemical quantity of sodium (Na) = 0.009 moles

Using the stoichiometry of the reaction, we can calculate the volume of hydrogen gas produced.

The molar volume of a gas at standard temperature and pressure (STP) is 22.4 L/mol.

Calculation:

The molar ratio of ethanol to hydrogen gas in the balanced chemical equation is 1:1. This means that for every mole of ethanol reacted, 1 mole of hydrogen gas is produced.

Given that the chemical quantity of ethanol is 0.5 moles, the volume of hydrogen gas produced can be calculated as follows: Volume of hydrogen gas (V(H2)) = 0.5 moles * 22.4 L/mol Volume of hydrogen gas (V(H2)) = 11.2 L

Therefore, the volume of hydrogen gas produced, when 0.5 moles of ethanol and 0.009 moles of sodium react, is 11.2 liters.

I hope this helps! If you have further questions or need additional assistance, feel free to ask.