PLS HELPP ...All alkali metals react with water to produce hydrogen gas and the corresponding alkali metal hydroxide A typical is that between lithium and water: 2Li(s) + 2H2O(l)  2LiOH(aq) + H2(g)

(a) How many moles of H2 can be formed by the complete reaction of 6.23 moles of Li with

water?

(b) How many grams of H2 can be formed by the complete reaction of 80.57 g of Li with

water?

A) 3.115 mol

B) 11.60 g h2

Explanation:

From the equation

2 mol Li gives 1 mol H2

Then 6.23 gives??

6.23 /2 = 3.116 mol H2

B) moles of Li =mass! molar mass

= 80.57/ 6.941 = 11.6 mol Li

Then moles of H2 = 11.6 /2 = 5.8 mol H2

Mass of H2 = mol * molar mass

5.8*2 =11.6 g h2

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:

some sources quote the lethal dose, for an average 170lbs opiate-naive individual is between 75 and 375 mg