Respuesta :

Answer:

125.08 grams

Explanation:

To determine the amount of SO3 produced, we need to use stoichiometry, which is the study of the quantitative relationships between reactants and products in a chemical reaction.  First, let's calculate the molar mass of O2. Oxygen (O) has an atomic mass of approximately 16 g/mol. Since O2 consists of two oxygen atoms, the molar mass of O2 is 2 * 16 g/mol = 32 g/mol.  Next, we need to find the number of moles of O2. We can do this by dividing the given mass (25.0 g) by the molar mass of O2 (32 g/mol):  25.0 g O2 / 32 g/mol = 0.78125 mol O2  According to the balanced equation, the ratio between O2 and SO3 is 1:2. This means that for every 1 mole of O2, 2 moles of SO3 are produced.  Since we have 0.78125 moles of O2, we can use the ratio to find the number of moles of SO3:  0.78125 mol O2 * (2 mol SO3 / 1 mol O2) = 1.5625 mol SO3  Finally, to calculate the mass of SO3, we need to multiply the number of moles of SO3 by its molar mass. The molar mass of SO3 can be calculated by adding the atomic masses of sulfur (S) and oxygen (O):   (32.07 g/mol S) + (3 * 16.00 g/mol O) = 80.07 g/mol SO3  Now we can calculate the mass of SO3:  1.5625 mol SO3 * 80.07 g/mol SO3 = 125.08 g SO3  Therefore, if there is 25.0 g of O2, approximately 125.08 grams of SO3 will be produced.

msm555

Answer:

125 grams

Explanation:

To determine the amount of SO₃ produced, we can use stoichiometry, which involves the molar ratios between the reactants and products in a balanced chemical equation.

The balanced equation for the reaction is:

[tex] 2 \text{SO}_2 + \text{O}_2 \longrightarrow 2 \text{SO}_3 [/tex]

This equation implies that 1 mole of O₂ reacts to produce 2 moles of SO₃.

Convert the given mass of O₂ to moles using its molar mass.

[tex] \text{Molar mass of O}_2 = 32.00 \, \text{g/mol} [/tex]

[tex] \text{Moles of O}_2 = \dfrac{\text{Mass of O}_2}{\text{Molar mass of O}_2} [/tex]

[tex] \text{Moles of O}_2 = \dfrac{25.0 \, \text{g}}{32.00 \, \text{g/mol}} [/tex]

[tex] \text{Moles of O}_2 \approx 0.78125 \, \text{mol} [/tex]

Use the stoichiometric coefficients to find the moles of SO₃ produced.

Since the balanced equation indicates a 1:2 ratio between moles of O₂ and moles of SO₃, we can say:

[tex] \text{Moles of SO}_3 = 2 \times \text{Moles of O}_2 [/tex]

[tex] \text{Moles of SO}_3 = 2 \times 0.78125 \, \text{mol} [/tex]

[tex] \text{Moles of SO}_3 = 1.5625 \, \text{mol} [/tex]

Convert moles of SO₃ to grams using its molar mass.

[tex] \text{Molar mass of SO}_3 = 80.06 \, \text{g/mol} [/tex]

[tex] \text{Mass of SO}_3 = \text{Moles of SO}_3 \times \text{Molar mass of SO}_3 [/tex]

[tex] \text{Mass of SO}_3 = 1.5625 \, \text{mol} \times 80.06 \, \text{g/mol} [/tex]

[tex] \text{Mass of SO}_3 \approx 125.09375\text{g} [/tex]

[tex] \text{Mass of SO}_3 \approx 125 \, \text{g (in nearest whole number)} [/tex]

Therefore, if 25.0 g of O₂ is reacted, approximately 125 grams of SO₃ will be produced.

Otras preguntas