What is the percent yield if 108.40 g O2 reacts with excess NH3 according to the balanced equation below and the actual yield is found to be 203.50 grams of H2O? 4 NH3 + 5 O2 → 4 NO + 6 H2O

To express the efficiency of a reaction, you can calculate the percent yield using this formula: %yield = (actual yield/theoretical yield) x 100. A percent yield of 90% means the reaction was 90% efficient, and 10% of the materials were wasted (they failed to react, or their products were not captured).

Explanation:

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:

the millers tale is the answer.