How much 16M nitric acid would react with 5.50 g of KOH?

Question

according to this equation: HNO3 KOH=H20 KNO3

Looking for any help

Answer 1

Calculate moles of KOH, use the balanced equation to relate that to moles HNO3, and finally L of 16 M HNO3.

5.50 g KOH x (1 mole KOH / 56.1 g KOH)

x (1 mole HNO3 / 1 mole KOH)

x (1 L HNO3 / 16 moles HNO3) = 0.0061 L

= 6.1 mL 16 M HNO3

Answer 2

Acid-base reactions are ubiquitous. In aqueous solutions acids increase the hydrogen ion (H+) concentration. On the other hand bases increase the hydroxide ion (OH-) concentration. When an acid and a base react in an aqueous solution the H+ and OH- ions combine to form water. These ions thus "neutralize" one another:

    

  H  + OH-  = H2O

                              

Answer 3

HNO3 + KOH => H20 + KNO3

One mole of HNO3 reacts completely with one mole of KOH

HNO3: 63

KOH:  57

5.5g KOH = 5.5/57 =  0.0965 moles

0.0965 moles of HNO3 will be provided by 0.0965*1/16 = 0.006031L = 6.031 ml of 16M HNO3

6.031 ml is the answer

Answer 4

I will answer to your Q, with another one. What do u intend to blow up with the resulting stuff?