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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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Q: How many moles of base are needed to neutralize 2 mol of acid for HNO3 plus KOH?
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How many moles of potassium hydroxide are needed to neutralize 3 moles of nitric acid?

3 moles KOH to neutralize 3 moles HNO3 : 3 mol OH- will react with 3 mol H+


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

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

20 moles of NaOH needed to neutralize 20 moles of nitric acid


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 )


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

4 moles HNO3 (63.018 grams/1 mole HNO3) = 252 grams nitric acid ================


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

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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


What is the pH of a solution that contains 1.32 grams of nitric acid dissolved in 750 milliters of water?

Two steps. Find molarity of nitric acid and need moles HNO3.Then find pH. 1.32 grams HNO3 (1 mole HNO3/63.018 grams) = 0.020946 moles nitric acid ------------------------------------- Molarity = moles of solute/Liters of solution ( 750 milliliters = 0.750 Liters ) Molarity = 0.020946 moles HNO3/0.750 Liters = 0.027928 M HNO3 ----------------------------------finally, - log(0.027928 M HNO3) = 1.55 pH ==========( could call it 1.6 pH )


How many moles of no2 will be obtained starting with 0.4 moles of nitric acid?

Starting with the formula: 2HNO3 --> H2O + NO2 If you have 0.4 moles of nitric acid (HNO3), you will get half the number of moles of NO2. So, you will have 0.2 moles of nitric acid.


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?

Make sure that the equation is balanced. NaOH + HNO3 > > NaNO3 + H2O ( all one to one ) 15.0 grams NaOH (1mol NaOH=39.998g) (1mol HNO3/1mol NaOH) = 0.375 moles of HNO3 Molarity = mols solute/volume solution 2.00M HNO3 = 0.375 mols/X volume = 0.188 Liters or, 188 milliliters.


How many moles of NO2 will be obtained strating with 0.4 moles of nitric acid?

Starting with the formula: 2HNO3 --> H2O + NO2 If you have 0.4 moles of nitric acid (HNO3), you will end up with half the moles of nitrogen dioxide (NO2)...so you will have 0.2 moles.


How many moles of hno3 are present in 40.0 mL of a 1.80 M solution of nitric acid?

7