Sulfur and oxygen form both sulfur dioxide and sulfur trioxide. when samples of these were decomposed the sulfur dioxide produced 3.49 g oxygen and 3.50 g sulfur, while the sulfur trioxide produced 6.75 g oxygen and 4.50 g sulfur. calculate the mass of oxygen per gram of sulfur for each sample and show that these results are consistent with the law of multiple proportions.

The reaction for decomposition of sulfur dioxide (SO₂) and sulfur trioxide (SO₃) are given below:

SO₂ → S + O₂    and  SO₃→ S + O₂

In SO₂, for 3.50 g sulfur,  oxygen produced is equal to 3.49 g, so, for 1 g sulfur 3.49/3.50 g= 0.997 g of O₂

In SO₃, for 4.50 g sulfur, oxygen produced is 6.75 g, so for 1 g sulfur, oxygen p;roduced will be equal to  6.75/4.50g=1.5 g of O₂.

With 1 g of sulfur, the ratio of the amount of oxygen present in SO₂ and SO₃ is 0.997:1.50= 2:3, this is a simple ratio and small whole number. These results are consistent with the Law of Multiple Proportion.

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: