Question

Решите пж СРОЧНО При сгорании 24 г магния образовалось 35 г его оксида. Определить выход продукта от теоретически возможного.

Answer 1

m(Mg) = 24 g

m(MgO практического) = 35 g

η(MgO) - ?

n(Mg) = 24 g / 24 g / mol = 1 mol

2Mg + O2 = 2MgO

2n(Mg) = 2n(MgO) = 1 mol;

m(MgO теоретического) = 1 mol * 40 g / mol = 40 g ;

η(MgO) = 35 g / 40 g = 0.875 = 87.5% ;

Ответ: 87.5% .

Answer 2

Task: Finding the yield of magnesium oxide

To determine the yield of a chemical reaction, we need to compare the actual amount of product obtained to the theoretical amount that should have been produced based on stoichiometry. In this case, we have 24 grams of magnesium (Mg) that reacted to form 35 grams of magnesium oxide (MgO). We want to calculate the yield of the magnesium oxide.

To find the theoretical yield of magnesium oxide, we need to consider the balanced chemical equation for the reaction between magnesium and oxygen:

2Mg + O2 → 2MgO

According to the balanced equation, 2 moles of magnesium react with 1 mole of oxygen to produce 2 moles of magnesium oxide. The molar mass of magnesium is approximately 24.31 g/mol, and the molar mass of oxygen is approximately 16.00 g/mol.

To calculate the theoretical yield, we can follow these steps:

1. Convert the mass of magnesium to moles using its molar mass: - Moles of Mg = Mass of Mg / Molar mass of Mg - Moles of Mg = 24 g / 24.31 g/mol

2. Use the stoichiometry of the balanced equation to determine the moles of magnesium oxide that should have been produced: - Moles of MgO = Moles of Mg * (2 moles of MgO / 2 moles of Mg)

3. Convert the moles of magnesium oxide to grams using its molar mass: - Mass of MgO = Moles of MgO * Molar mass of MgO - Mass of MgO = Moles of MgO * (24.31 g/mol + 16.00 g/mol)

4. Calculate the percentage yield by dividing the actual yield (35 g) by the theoretical yield and multiplying by 100: - Percentage yield = (Actual yield / Theoretical yield) * 100

Let's calculate the theoretical yield of magnesium oxide using the given information:

1. Moles of Mg = 24 g / 24.31 g/mol = 0.987 moles of Mg 2. Moles of MgO = 0.987 moles of Mg * (2 moles of MgO / 2 moles of Mg) = 0.987 moles of MgO 3. Mass of MgO = 0.987 moles of MgO * (24.31 g/mol + 16.00 g/mol) = 35.01 g of MgO (theoretical yield)

Now we can calculate the percentage yield:

Percentage yield = (35 g / 35.01 g) * 100 = 99.97%

Therefore, the yield of magnesium oxide is approximately 99.97% of the theoretical yield.

Please note that the calculated values are approximate due to rounding errors in the molar masses and the assumption of ideal conditions.