You are provided with 6.05 g s o 2 to react with 2.23 g o 2 . indicate your theoretical yield.

7.56 g

Explanation:

Data: Molar masses

SO₂: 64.07 g/mol

O₂: 32.00 g/mol

SO₃: 80.07 g/mol

Let's consider the following balanced reaction.

SO₂ + 1/2 O₂ → SO₃

The theoretical mass ratio of SO₂ to O₂ is 64.07g to 1/2 × 32.00 g = 16.00 g, i.e. 4.004 g SO₂/ 1 g O₂.

The experimental mass ratio of SO₂ to O₂ is 6.05 g to 2.23 g, i.e. 2.713 g SO₂/ 1 g O₂.

Comparing both ratios, we can determine that SO₂ is the limiting reactant.

The mass ratio of SO₂ to SO₃ is 64.07 g to 80.07 g. The theoretical yield of SO₃ from 6.05 g of SO₂ is:

6.05 g SO₂ × (80.07 g SO₃/ 64.07 g SO₂) = 7.56 g SO₃

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: