How many mL of a 0.4 M nitric acid solution are required to neutralize 200 mL of a 0.16 M potassium hydroxide

80 ml.

Explanation:

The balanced equation is:

KOH + HNO3 >  KNO3 + H2O

So 1 mole of KOH reacts with 1 mole of HNO3

200ml of 0.16M KOH contains 0.16 * 200/1000

= 0.16 * 0.2

= 0.032 M KOH.

So we need the volume (V) of 0.4 M HNO3 which contains 0.032 moles.

0.4 * V = 0.032

V = 0.032 / 0.4

V = 0.08 liters

= 80 mls.

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: