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Since both the acid and the base have equivalent weights equal to their formula weights, 2 moles of KOH are needed to neutralize 2 moles of nitric acid.

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How many moles of sodium hydroxide are required to neutralize 20 mol of nitric acid?

The balanced chemical equation for the neutralization between sodium hydroxide (NaOH) and nitric acid (HNO3) is 1 mol of NaOH reacts with 1 mol of HNO3. Therefore, 20 moles of nitric acid would require 20 moles of sodium hydroxide to neutralize it.


How many moles of potassium hydroxide are needed to neutralize 3 moles of nitric acid?

The balanced chemical equation for the reaction between potassium hydroxide (KOH) and nitric acid (HNO3) is 1:1 ratio. Therefore, 3 moles of nitric acid will require 3 moles of potassium hydroxide to neutralize it.


A nitric acid soulution is neutralized using sodium hydroxide how many grams of sodium hydroxide are needed to neutralize 2.50 L of 0.800 M Nitriric acid soulution?

Balanced equation first. HNO3 + NaOH >> NaNO3 + H2O Now, Molarity = moles of solute/volume of solution ( find moles HNO3 ) 0.800 M HNO3 = moles/2.50 Liters = 2 moles of HNO3 ( these reactants are one to one, so we can proceed to grams NaOH ) 2 moles NaOH (39.998 grams NaOH/1mole NaOH) = 79.996 grams of NaOH need to neutralize the acid. ( you do significant figures )


Lakes have been acidified by acid rain hno3 and h2so4 can be neutralized by a process called liming in which limestone caco3 is added to the acudified water what mass of limestone in kg would?

To calculate the mass of limestone (CaCO3) needed to neutralize HNO3 and H2SO4 in acidified water, you can use stoichiometry. First, determine the amount of acid in moles. Then, use the balanced chemical equation to find the moles of CaCO3 needed to neutralize the acid. Finally, convert the moles of CaCO3 to mass in kilograms using its molar mass.


How many moles of the base are required to neutralize one mole of the acid for the reaction of HNO3 and KOH?

For the reaction between HNO3 (acid) and KOH (base), it is a 1:1 molar ratio reaction. This means that 1 mole of HNO3 will react with 1 mole of KOH. So, 1 mole of KOH is required to neutralize 1 mole of HNO3 in this reaction.


How many moles of HNO3 are present if 0.129 mol of Ba(OH)2 was needed to neutralize the acid solution?

2


A nitric acid solution is neutralized using sodium hydroxide How many grams of sodium hydroxide are needed to neutralize 2.50 L of 0.800 M nitric acid solution?

To neutralize the nitric acid, you need a 1:1 mole ratio of sodium hydroxide to nitric acid. First, calculate the moles of nitric acid in the solution using the formula Molarity = moles/volume. Then, use the mole ratio to find the moles of sodium hydroxide needed. Finally, convert this to grams using the molar mass of sodium hydroxide.


What is the molarity of a nitric acid solution HNO3 if 20.0 mL of the solution is needed to exactly neutralize 10.0 mL of a 1.67 M NaOH solution?

The balanced chemical equation for the neutralization reaction is: HNO3 + NaOH → NaNO3 + H2O From the equation, we know that the mole ratio of HNO3 to NaOH is 1:1. This means that the moles of HNO3 will be equal to the moles of NaOH. First, calculate the moles of NaOH used: Moles NaOH = (10.0 mL) x (0.001 L/mL) x (1.67 mol/L) = 0.0167 mol Since the moles of NaOH are equal to the moles of HNO3, we have 0.0167 mol of HNO3 in 20.0 mL of solution: Molarity of HNO3 = moles HNO3 / volume of solution (L) = 0.0167 mol / 0.020 L = 0.835 M


How many times do you have to neutralizer bases using an acid?

The number of moles of acid and the base should be equal to neutralize. (So the number of moles of base is needed to answer this correctly)


How many grams would be in four 4 moles of nitric acid HNO3?

The molar mass of nitric acid (HNO3) is 63.01 g/mol. To find the total grams in 4 moles, you would multiply the molar mass by the number of moles: 63.01 g/mol x 4 mol = 252.04 grams. So, there would be 252.04 grams in four moles of nitric acid (HNO3).


What is the molarity of a solution dissolve 0.31 grams of HNO3 in 300ml of water?

Molarity = moles of solute/Liters of solution ( get moles of HNO3 and 300 ml = 0.300 Liters ) 0.31 grams Nitric acid (1 mole HNO3/63.018 grams) = 0.004919 moles HNO3 Molarity = 0.004919 moles HNO3/0.300 Liters = 0.0164 M HNO3


How many moles of HNO3 are required to react with 6 moles of Cu?

The balanced chemical equation for the reaction between copper (Cu) and nitric acid (HNO3) is: 3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O According to the equation, 8 moles of HNO3 are required to react with 3 moles of Cu. Therefore, to react with 6 moles of Cu, you would need 16 moles of HNO3.