How many liters of nitric acid are needed to neutralize 25mL of 4.2M calcium hydroxide?

the reaction is:

2HNO3 + Ca(OH)2 --> Ca(NO3)2 + 2H2O

we know the mole of calcium hydroxide
n=cv = 4.2mol/L x 0.025L = _____

this mole needs to be neutralized by the acid. so we need to calculate the mole of nitric acid using the mole ratio from the balanced equation;

2 mole of nitric acid neutralize 1 mole of calcium hydroxide.

2:1
x:____(the mole of calcium hydroxide)

where x is the mole of nitric acid. use that mole to calculate the volume..can you???..we need a concentration of nitric acid...is there a concentration for nitric acid??? check again..

41ml

To determine the amount of nitric acid needed to neutralize calcium hydroxide, we need to use stoichiometry. The balanced chemical equation for the reaction between nitric acid (HNO3) and calcium hydroxide (Ca(OH)2) is:

2 HNO3 + Ca(OH)2 -> Ca(NO3)2 + 2 H2O

From the equation, we can see that it takes 2 moles of nitric acid to react with 1 mole of calcium hydroxide.

First, we need to calculate the number of moles of calcium hydroxide in 25 mL of a 4.2M solution:

N = C x V
N = 4.2 mol/L x 0.025 L
N ≈ 0.105 moles

Since the stoichiometric ratio is 2:1, we can now determine the number of moles of nitric acid required:

Number of moles of nitric acid = 0.105 moles x 2
Number of moles of nitric acid ≈ 0.21 moles

Finally, to convert moles to liters, we can use the formula:

V = N / C
V = 0.21 moles / 4.2 mol/L
V ≈ 0.05 L

Therefore, approximately 0.05 liters (or 50 mL) of nitric acid are needed to neutralize 25 mL of 4.2M calcium hydroxide.