Question

11. Напишите реакции до конца, расставьте коэффициенты (2 балла): А) KOH+CO2= Б) Ba(OH)2+HNO3= 12. Каковы оббьем (н.у) и масса водорода , выделившегося при взаимодействии 6,9 г. металла натрия с водой, 2Na + 2 H2O= 2 NaOH + H2 ?​

Answer 1

Ответ:

11. 2KOH + CO2 → K2CO3 + H2O

Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O

Answer 2

Reaction Equations and Coefficients

A) The balanced chemical equation for the reaction between KOH and CO2 is: 2 KOH + CO2 = K2CO3 + H2O

B) The balanced chemical equation for the reaction between Ba(OH)2 and HNO3 is: Ba(OH)2 + 2 HNO3 = Ba(NO3)2 + 2 H2O

Volume and Mass of Hydrogen Gas

The reaction between sodium and water can be represented by the balanced chemical equation: 2 Na + 2 H2O = 2 NaOH + H2

Given that 6.9 g of sodium reacts with water, we can calculate the volume and mass of hydrogen gas produced using the stoichiometry of the reaction.

First, we need to determine the moles of sodium involved in the reaction: Molar mass of sodium (Na) = 22.99 g/mol Moles of sodium = Mass of sodium / Molar mass of sodium Moles of sodium = 6.9 g / 22.99 g/mol

Next, we use the stoichiometry of the reaction to find the moles of hydrogen gas produced: From the balanced equation, 2 moles of sodium produce 1 mole of hydrogen gas. Moles of hydrogen gas = Moles of sodium / 2

Finally, we can calculate the volume and mass of hydrogen gas using the ideal gas law: Volume of hydrogen gas at standard temperature and pressure (STP) = Moles of hydrogen gas * 22.4 L/mol Mass of hydrogen gas = Moles of hydrogen gas * molar mass of hydrogen gas

By performing these calculations, we can determine the volume and mass of hydrogen gas produced in the reaction.

I hope this helps! If you have any further questions, feel free to ask.