Question

Яка маса метану утвориться при взаємодії 200г алюміній карбіду з достатньою кількістю води?

Answer 1

Объяснение:

АС; + 12Н>О + ЗСН, + 4А(ОН); Молярная масса М(А!.С.) = 4*27+3*12 = 144 г/моль,

224 мл СН, = 0,224 л СН..

С учетом потерь нужно получить метана \(СН.,) = 0,224 л * 100% / 96% = 0,233 л.

По уравнению реакции из 144 г/моль А.С. образуется 3*22,4 л СН.. ИзхгА|4С. получится 0,233 л СН..

х = 144 г/моль * 0,233 л /(3*22,4 л/моль) = 0,499 ГАЦС..

Потребуется 0,499 г А\.С..

Answer 2

Mass of Methane Formed from the Reaction

When 200g of aluminum carbide reacts with a sufficient amount of water, it forms methane. The mass of methane formed can be calculated using stoichiometry and the molar masses of the compounds involved.

The balanced chemical equation for the reaction is:

3Al4C3 + 12H2O → 3CH4 + 4Al(OH)3

From the balanced equation, it's clear that 3 moles of methane are formed for every 1 mole of aluminum carbide.

The molar mass of aluminum carbide (Al4C3) is approximately 144 g/mol, and the molar mass of methane (CH4) is approximately 16 g/mol.

Using this information, we can calculate the mass of methane formed:

1. Calculate the moles of aluminum carbide: - Moles of Al4C3 = Mass of Al4C3 / Molar mass of Al4C3 - Moles of Al4C3 = 200g / 144 g/mol ≈ 1.39 moles

2. Use the mole ratio from the balanced equation to find moles of methane: - Moles of CH4 = Moles of Al4C3 * (3 moles of CH4 / 1 mole of Al4C3) - Moles of CH4 ≈ 1.39 moles * 3 ≈ 4.17 moles

3. Calculate the mass of methane formed: - Mass of CH4 = Moles of CH4 * Molar mass of CH4 - Mass of CH4 ≈ 4.17 moles * 16 g/mol ≈ 66.7g

Therefore, approximately 66.7g of methane will be formed when 200g of aluminum carbide reacts with a sufficient amount of water