Question

До міді додали концентрований розчин нітратної кислоти. Мідь прореагувала з кислотою з утворенням синього розчину купрум(II) нітрату та виділенням буро го газу нітроген(IV) оксиду та води. Склади рівняння цієї окисно-відновної реакції. Добери коефіцієнти, застосувавши метод електронного балансу. Обчисли масу міді, потрібну для одержання купрум(ІІ) нітрату кількістю речовини 3 моль.

Answer 1

Ответ:

На фотографии…

Answer 2

Reaction Equation and Coefficients

The reaction between copper and concentrated nitric acid can be represented by the following balanced equation:

Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O [[1]]

In this equation, copper (Cu) reacts with concentrated nitric acid (HNO3) to form copper(II) nitrate (Cu(NO3)2), nitrogen dioxide gas (NO2), and water (H2O).

The coefficients in the balanced equation indicate the stoichiometric ratios between the reactants and products. In this case, one mole of copper reacts with four moles of nitric acid to produce one mole of copper(II) nitrate, two moles of nitrogen dioxide gas, and two moles of water.

Calculation of Copper Mass

To calculate the mass of copper required to obtain 3 moles of copper(II) nitrate, we need to use the molar mass of copper and the stoichiometric ratio from the balanced equation.

The molar mass of copper (Cu) is approximately 63.55 g/mol [[2]]. According to the balanced equation, one mole of copper reacts to produce one mole of copper(II) nitrate. Therefore, the molar mass of copper(II) nitrate (Cu(NO3)2) is:

2 * (63.55 g/mol) + 2 * (14.01 g/mol) + 6 * (16.00 g/mol) = 187.55 g/mol

To find the mass of copper required to obtain 3 moles of copper(II) nitrate, we can use the following calculation:

Mass of copper = (3 moles of copper(II) nitrate) * (63.55 g/mol) / (1 mole of copper(II) nitrate) Mass of copper = 190.65 g

Therefore, approximately 190.65 grams of copper are required to obtain 3 moles of copper(II) nitrate.

I hope this helps! Let me know if you have any further questions.