Question

При взаимодействии железа с хлором образуется хлорид железа ( III ). Рассчитайте количество вещества, массу и объём ( н.у. ) хлора, который потребуется для взаимодействия с 280 г железа, содержащего 20% примесей?

Answer 1

Ответ:

134,4 л, 6 моль, 426 г

Объяснение:

в фото

Answer 2

Calculation of the Amount of Chlorine Required

To calculate the amount of chlorine required for the reaction with 280 g of iron, we need to consider the stoichiometry of the reaction and the percentage of impurities in the iron.

The balanced chemical equation for the reaction between iron and chlorine is:

2 Fe + 3 Cl2 -> 2 FeCl3

From the equation, we can see that 2 moles of iron react with 3 moles of chlorine to produce 2 moles of iron(III) chloride.

To calculate the amount of chlorine required, we can use the following steps:

Step 1: Calculate the moles of iron in 280 g of iron. - The molar mass of iron (Fe) is approximately 55.85 g/mol. - The moles of iron can be calculated using the formula: moles = mass / molar mass. - Substituting the values, we get: moles of iron = 280 g / 55.85 g/mol.

Step 2: Calculate the moles of iron(III) chloride produced. - From the balanced equation, we know that 2 moles of iron react with 3 moles of chlorine to produce 2 moles of iron(III) chloride. - Since the stoichiometry is 2:3, the moles of iron(III) chloride produced will be the same as the moles of chlorine required.

Step 3: Calculate the moles of chlorine required. - The moles of chlorine required can be calculated using the formula: moles = (moles of iron / 2) * 3.

Step 4: Calculate the mass of chlorine required. - The mass of chlorine required can be calculated using the formula: mass = moles * molar mass of chlorine. - The molar mass of chlorine (Cl2) is approximately 70.90 g/mol.

Step 5: Calculate the volume of chlorine required at standard temperature and pressure (STP). - The volume of chlorine required can be calculated using the formula: volume = moles * molar volume at STP. - The molar volume at STP is approximately 22.4 L/mol.

Let's perform the calculations:

Step 1: Calculate the moles of iron in 280 g of iron. - Moles of iron = 280 g / 55.85 g/mol = 5.01 mol.

Step 2: Calculate the moles of iron(III) chloride produced. - Moles of iron(III) chloride = Moles of iron = 5.01 mol.

Step 3: Calculate the moles of chlorine required. - Moles of chlorine = (Moles of iron / 2) * 3 = (5.01 mol / 2) * 3 = 7.52 mol.

Step 4: Calculate the mass of chlorine required. - Mass of chlorine = Moles of chlorine * Molar mass of chlorine = 7.52 mol * 70.90 g/mol = 532.89 g.

Step 5: Calculate the volume of chlorine required at STP. - Volume of chlorine = Moles of chlorine * Molar volume at STP = 7.52 mol * 22.4 L/mol = 168.45 L.

Therefore, the amount of substance (moles) of chlorine required is approximately 7.52 mol, the mass of chlorine required is approximately 532.89 g, and the volume of chlorine required at STP is approximately 168.45 L.

Please note that these calculations assume ideal conditions and may vary in practice due to factors such as reaction efficiency and impurities in the reactants.

Let me know if there's anything else I can help you with!