Question

При взаимодействии SO3 массой 40г с H2O массой 10г образовалось H2SO4 (серная кислота). Найти массу H2SO4. Заранее спасибо за ответ!

Answer 1

H2O + SO3 = H2SO4

n(SO3)=40/80=0.5 моль

n(H2O)=10/18=0,56 моль

На 1 часть SO3 должна приходиться 1 часть H2O, а тут больше. Следовательно, вода в избытке.

Считаем по SO3...

 

n(H2SO4)=n(SO3)=0.5 моль

m(H2SO4)=0.5 моль * 98 г/моль = 49 г

 

Answer 2

SO3+H2O=H2SO4

n SO3=40/80=0.5 моль

 

n H2O=10/18=0.5555 моль

n H2SO4=0.5 моль

m H2SO4=0.5*98=49 г

 

 

 

 

 

 

Answer 3

Calculation of the Mass of H2SO4 Formed

To find the mass of H2SO4 formed when 40g of SO3 reacts with 10g of H2O, we need to determine the stoichiometry of the reaction and use it to calculate the mass of the product.

The balanced chemical equation for the reaction between SO3 and H2O to form H2SO4 is:

SO3 + H2O → H2SO4

From the equation, we can see that the molar ratio between SO3 and H2SO4 is 1:1. This means that for every 1 mole of SO3 reacted, 1 mole of H2SO4 is formed.

To calculate the mass of H2SO4 formed, we need to convert the given masses of SO3 and H2O to moles using their respective molar masses. The molar mass of SO3 is 80.06 g/mol, and the molar mass of H2O is 18.02 g/mol.

1. Calculate the number of moles of SO3: - Mass of SO3 = 40g - Molar mass of SO3 = 80.06 g/mol - Moles of SO3 = Mass of SO3 / Molar mass of SO3

2. Calculate the number of moles of H2O: - Mass of H2O = 10g - Molar mass of H2O = 18.02 g/mol - Moles of H2O = Mass of H2O / Molar mass of H2O

3. Determine the limiting reactant: - The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed. - To find the limiting reactant, compare the moles of each reactant and identify the one with the smaller value. - The reactant with the smaller number of moles is the limiting reactant.

4. Calculate the moles of H2SO4 formed: - Since the stoichiometry of the reaction is 1:1 between SO3 and H2SO4, the moles of H2SO4 formed will be equal to the moles of the limiting reactant.

5. Calculate the mass of H2SO4 formed: - Mass of H2SO4 = Moles of H2SO4 formed * Molar mass of H2SO4

Let's perform the calculations now.

Calculation:

1. Moles of SO3: - Mass of SO3 = 40g - Molar mass of SO3 = 80.06 g/mol - Moles of SO3 = 40g / 80.06 g/mol = 0.4999 mol (approximately 0.5 mol) 2. Moles of H2O: - Mass of H2O = 10g - Molar mass of H2O = 18.02 g/mol - Moles of H2O = 10g / 18.02 g/mol = 0.5558 mol (approximately 0.56 mol) 3. Determining the limiting reactant: - Comparing the moles of SO3 and H2O, we can see that the moles of SO3 (0.5 mol) are smaller than the moles of H2O (0.56 mol). - Therefore, SO3 is the limiting reactant.

4. Moles of H2SO4 formed: - Since the stoichiometry of the reaction is 1:1 between SO3 and H2SO4, the moles of H2SO4 formed will be equal to the moles of the limiting reactant, which is 0.5 mol.

5. Mass of H2SO4 formed: - Molar mass of H2SO4 = 98.09 g/mol - Mass of H2SO4 = Moles of H2SO4 formed * Molar mass of H2SO4 = 0.5 mol * 98.09 g/mol = 49.045 g (approximately 49.05 g)

Therefore, the mass of H2SO4 formed when 40g of SO3 reacts with 10g of H2O is approximately 49.05 grams.

Please note that the calculated mass is an approximation due to rounding off the number of moles.