Consider this reaction, which occurs in the atmosphere and contributes to photochemical smog: SO2(g) + H2O(l) →H2SO3(g) If there is 18.72 g SO2 and excess H2O present, the reaction yields 17.3 g H2SO3. Calculate the percent yield for the reaction.

72.2 % is the percent yield

Explanation:

Percent yield of a reaction is:

(Produced yield / Theoretical yield) . 100

The produced yield is 17.3 g of sulfurous acid.

Let's determine the theoretical yield. The reaction is:

SO₂(g) + H₂O(l) → H₂SO₃(g)

We convert the mass of sulfur dioxide to moles:

18.72 g / 64.06 g/mol = 0.292 moles.

As ratio is 1:1, for 0.292 moles of dioxide we can produce 0.292 moles of acid. Let's convert the moles to mass in order to determine the 100 % yield reaction. (Theoretical yield). 0.292 mol . 82.06 g/ 1 mol = 23.9 g

We replace data  → (17.3 g / 23.9 g) . 100 = 72.2%

The percent yield of the reactions is 72.1 %

Explanation:

Step 1: Data given

Mass of SO2 = 18.72 grams

Mass of H2SO3 produced = 17.3 grams

Molar mass of SO2 = 64.07 g/mol

Step 2: The balanced equation

SO2(g) + H2O(l) →H2SO3(g)

Step 3: Calculate moles SO2

Moles SO2 = mass SO2 / molar mass SO2

Moles SO2 = 18.72 grams / 64.07 g/mol

Moles SO2 = 0.2922 moles

Step 4: Calculate moles H2SO3

For 1 mol SO2 we need 1 mol H2O to produce 1 mol H2SO3

For 0.2922 moles SO2 we'll have 0.2922 moles H2SO3

Step 5: Calculate mass H2SO3

Mass H2SO3 = 0.2922 moles * 82.07 g/mol

Mass H2SO3 = 23.98 grams

Step 6: Calculate the percent yield

Percent yield = (actual mass/  theoretical mass) * 100%

Percent yield = ( 17.3 grams / 23.98 grams) * 100%

Percent yield = 72.1 %

The percent yield of the reactions is 72.1 %

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.