If you had excess chlorine, how many moles of of aluminum chloride could be produced from 11.0 g of aluminum

0.407 moles of aluminum chloride could be produced from 11.0 g of aluminum.

Explanation:

To calculate the moles :

According to stoichiometry :

is the limiting reagent as it limits the formation of product and is the excess reagent.

As 2 moles of give = 2 moles of

Thus moles of give =  of

Thus 0.407 moles of aluminum chloride could be produced from 11.0 g of aluminum.

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 answer for sodium is 0.186

the answer for rubidium is 0.251