Consider the neutralization reaction 2hno3(aq)+ba(oh)2(aq)⟶2h2o(l)+ba(no 3)2(aq) 2hno3(aq)+ba(oh)2(aq)⟶2h2o(l)+ba(no 3)2(aq) a 0.110 l0.110 l sample of an unknown hno3hno3 solution required 51.9 ml51.9 ml of 0.100 m ba(oh)20.100 m ba(oh)2 for complete neutralization. what is the concentration of the hno3hno3 solution? concentration:

The concentration of solution will be 0.094 M.

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

where,

are the n-factor, molarity and volume of acid which is

are the n-factor, molarity and volume of base which is

We are given:

Conversion factor: 1L = 1000 mL

Putting values in above equation, we get:

Hence, the concentration of solution will be 0.094 M.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

idk what the question is about