Question

при разложении 25 г известняка, согласно уравнению реакции CaCO3=> CaO + CO2 обр-сь 12 г оксида кальция.

Answer 1

Что надо найти в этой задачи ?

Answer 2

Decomposition of Limestone and Formation of Calcium Oxide and Carbon Dioxide

When 25 grams of limestone (CaCO3) decomposes according to the equation CaCO3 -> CaO + CO2, 12 grams of calcium oxide (CaO) and carbon dioxide (CO2) are formed.

The balanced chemical equation for the decomposition of limestone is: CaCO3 -> CaO + CO2

To find the amount of calcium oxide and carbon dioxide formed when 25 grams of limestone decomposes, we can use stoichiometry to calculate the quantities of the products.

First, we need to determine the molar mass of each substance involved: - The molar mass of CaCO3 (limestone) is approximately 100.09 g/mol. - The molar mass of CaO (calcium oxide) is approximately 56.08 g/mol. - The molar mass of CO2 (carbon dioxide) is approximately 44.01 g/mol.

Using the molar masses, we can calculate the number of moles of CaCO3 in 25 grams: moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3 moles of CaCO3 = 25 g / 100.09 g/mol

Now, we can use the balanced chemical equation to determine the moles of CaO and CO2 produced: 1 mole of CaCO3 produces 1 mole of CaO and 1 mole of CO2.

Therefore, the moles of CaO and CO2 produced will be the same as the moles of CaCO3.

Finally, we can calculate the mass of CaO and CO2 produced: mass of CaO = moles of CaO * molar mass of CaO mass of CO2 = moles of CO2 * molar mass of CO2

Substituting the values, we find: mass of CaO = moles of CaCO3 * molar mass of CaO mass of CaO = (25 g / 100.09 g/mol) * 56.08 g/mol mass of CaO ≈ 14 g

mass of CO2 = moles of CaCO3 * molar mass of CO2 mass of CO2 = (25 g / 100.09 g/mol) * 44.01 g/mol mass of CO2 ≈ 11 g

Therefore, when 25 grams of limestone decomposes, approximately 14 grams of calcium oxide and 11 grams of carbon dioxide are formed.

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