Question

ПОМОГИТЕ СРОЧНО ПОЖАЛУЙСТА!РЕШИТЕ ЗАДАЧУ!!СРОЧНО!!!!ПОЖАЛУЙСТА! При разложении малахита(CuOH)2Co3 образуются вода, оксид меди(II)и углекислый газ.Сколько грамм малахита разложили, если образовалось 0, 25 моль оксида меди(II)?

Answer 1

Оформление подобных задач учителя может отличаться от моего,но ответ верный

Answer 2

Problem Analysis

To solve this problem, we need to determine the amount of malachite (CuOH)2CO3 that was decomposed, given the formation of 0.25 moles of copper(II) oxide (CuO). We can use the balanced chemical equation for the decomposition reaction to calculate the amount of malachite.

Balanced Chemical Equation

The balanced chemical equation for the decomposition of malachite is as follows: 2(CuOH)2CO3 -> 2CuO + 2H2O + CO2

Calculation

According to the balanced equation, for every 2 moles of malachite decomposed, 2 moles of copper(II) oxide are formed. Therefore, the molar ratio between malachite and copper(II) oxide is 2:2, which simplifies to 1:1.

Given that 0.25 moles of copper(II) oxide were formed, we can conclude that the same amount of moles of malachite decomposed. Therefore, the amount of malachite decomposed is also 0.25 moles.

To calculate the mass of malachite decomposed, we need to use its molar mass. The molar mass of malachite can be calculated by summing the atomic masses of its constituent elements: - Copper (Cu) has an atomic mass of 63.55 g/mol - Oxygen (O) has an atomic mass of 16.00 g/mol - Hydrogen (H) has an atomic mass of 1.01 g/mol

The molar mass of malachite is: 2(63.55 g/mol) + 2(1.01 g/mol) + 3(16.00 g/mol) = 221.38 g/mol

To find the mass of malachite decomposed, we can use the following equation: mass = moles * molar mass

Substituting the values, we get: mass = 0.25 moles * 221.38 g/mol = 55.35 g

Therefore, 55.35 grams of malachite were decomposed to form 0.25 moles of copper(II) oxide.

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