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What is the number of moles of NaOH required to neutralize 1 mole of H2SO4?
Well, darling, if we're talking about a 1:2 molar ratio between NaOH and H2SO4, then you'd need 2 moles of NaOH to neutralize 1 mole of H2SO4. It's all about those stoichiometry dance moves, honey. Just make sure you're not tripping over your chemical equations!
How grams of NaOH is required to nuetralize 100 ml of 1 M H2SO4?
Since H2SO4 is a diprotic acid, it will require twice the amount of NaOH to neutralize it. Therefore, molarity of NaOH should also be 1 M. 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, to neutralize 1 mole of H2SO4, 2 moles of NaOH are required. To neutralize 1 mole of H2SO4 in 100 ml (0.1 L) of 1 M solution, you will need 0.1 moles of NaOH.
What volume of 0.744 Molarity of NaOH is required to titrate 10.00ml of 0.526 M H2SO4?
The balanced chemical equation for the reaction between NaOH and H2SO4 is 2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O. From the equation, it is a 1:1 ratio of NaOH to H2SO4. Therefore, to neutralize 10.00 ml of 0.526 M H2SO4, you will need the same amount of 0.526 M NaOH, which is 10.00 ml.
How many grams of sulfuric acid will neutralize 10.0 g of sodium hydroxide?
For every mole of sodium hydroxide, you need 1 mole of sulfuric acid for neutralization. The molar mass of sodium hydroxide (NaOH) is 40.0 g/mol and sulfuric acid (H2SO4) is 98.1 g/mol. So, to neutralize 40 g of NaOH (1 mole), you would need 98.1 g of H2SO4 (1 mole). Therefore, to neutralize 10.0 g of NaOH, you would need 24.53 g of H2SO4.
How much 66 baume sulfuric acid is required to neutralize 350lbs of sodium hydroxide?
66 Baume Sulfuric Acid is contains 93.2% H2SO4 by weight in water. It requires 1 lbmol of H2SO4 to neutralize 2 lbmol of NaOH. Assuming that 350 lbs of NaOH is pure, there are 8.75 lbmol of NaOH. (Divide 350lb by 40 lb/lbmol, the MW of NaOH.) It takes 4.375 lbmol of H2SO4 to neutralize the 350 lb of NaOH because there are 2 lbmol of H per lbmol of H2SO4 and one lbmol of OH per lbmol of NaOH. 4.375 lbmol is equivalent to 428.75 lb of H2SO4. (Multiply by 98 lb/lbmol, the MW of H2SO4.) Then divide by 93.2% or 0.932 to get the quanty of 66 Baume containing 428.75 lb of H2SO4. 460 lb of 66 Baume. The density of 66 Baume is 1.8354 g/cm3 which is equivalent to 15.32 lb/gallon. (Mulitply by 8.345.) Finally, it requires 30 gallons of 66 baume to neutralize 350 lbs of NaOH. (Divide 460 by 15.32.)
What is the number of moles of NaOH required to neutralize 1 mole of H2SO4?
Well, darling, if we're talking about a 1:2 molar ratio between NaOH and H2SO4, then you'd need 2 moles of NaOH to neutralize 1 mole of H2SO4. It's all about those stoichiometry dance moves, honey. Just make sure you're not tripping over your chemical equations!
How grams of NaOH is required to nuetralize 100 ml of 1 M H2SO4?
Since H2SO4 is a diprotic acid, it will require twice the amount of NaOH to neutralize it. Therefore, molarity of NaOH should also be 1 M. 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, to neutralize 1 mole of H2SO4, 2 moles of NaOH are required. To neutralize 1 mole of H2SO4 in 100 ml (0.1 L) of 1 M solution, you will need 0.1 moles of NaOH.
What volume of 0.744 Molarity of NaOH is required to titrate 10.00ml of 0.526 M H2SO4?
The balanced chemical equation for the reaction between NaOH and H2SO4 is 2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O. From the equation, it is a 1:1 ratio of NaOH to H2SO4. Therefore, to neutralize 10.00 ml of 0.526 M H2SO4, you will need the same amount of 0.526 M NaOH, which is 10.00 ml.
How many grams of sulfuric acid will neutralize 10.0 g of sodium hydroxide?
For every mole of sodium hydroxide, you need 1 mole of sulfuric acid for neutralization. The molar mass of sodium hydroxide (NaOH) is 40.0 g/mol and sulfuric acid (H2SO4) is 98.1 g/mol. So, to neutralize 40 g of NaOH (1 mole), you would need 98.1 g of H2SO4 (1 mole). Therefore, to neutralize 10.0 g of NaOH, you would need 24.53 g of H2SO4.
How much 66 baume sulfuric acid is required to neutralize 350lbs of sodium hydroxide?
66 Baume Sulfuric Acid is contains 93.2% H2SO4 by weight in water. It requires 1 lbmol of H2SO4 to neutralize 2 lbmol of NaOH. Assuming that 350 lbs of NaOH is pure, there are 8.75 lbmol of NaOH. (Divide 350lb by 40 lb/lbmol, the MW of NaOH.) It takes 4.375 lbmol of H2SO4 to neutralize the 350 lb of NaOH because there are 2 lbmol of H per lbmol of H2SO4 and one lbmol of OH per lbmol of NaOH. 4.375 lbmol is equivalent to 428.75 lb of H2SO4. (Multiply by 98 lb/lbmol, the MW of H2SO4.) Then divide by 93.2% or 0.932 to get the quanty of 66 Baume containing 428.75 lb of H2SO4. 460 lb of 66 Baume. The density of 66 Baume is 1.8354 g/cm3 which is equivalent to 15.32 lb/gallon. (Mulitply by 8.345.) Finally, it requires 30 gallons of 66 baume to neutralize 350 lbs of NaOH. (Divide 460 by 15.32.)
What volume (in liters) of a 1.9 M NaOH solution would neutralize 7.2 moles of H2SO4 (diprotic)?
To neutralize 1 mole of diprotic acid (H2SO4), you need 2 moles of NaOH. Therefore, to neutralize 7.2 moles of H2SO4, you would need 14.4 moles of NaOH. Using the formula M = mol/L, where M is the molarity, mol is the amount of solute in moles, and L is the volume in liters, you can calculate the volume of the 1.9 M NaOH solution needed as 7.57 liters.
A 25.00ml sample of h2so4 requires 22.65 ml of the 0.550m naoh for its titration what was the concentration of sulfuric acid?
The balanced chemical equation for the reaction is H2SO4 + 2NaOH -> Na2SO4 + 2H2O. From the mole ratio, 1 mole of H2SO4 reacts with 2 moles of NaOH. Using the volume and concentration of NaOH, we can calculate the moles of NaOH used. Then, knowing the moles of NaOH used and the volume of H2SO4, we can find the concentration of sulfuric acid.
What volum of 0.502 m naoh solution would be required to neutralize 27.2 ml of 0.491 m hno solution?
The molar weight of NaOH is 39.9971 g/mol. .15g is 3.75 millimoles. The reaction of H2SO4 and NaOH is H2SO4 + 2 NaOH --> Na2SO4 + 2 H2O. So the molar ratio of H2SO4 to NaOH is 1:2. So it takes 1.825 millimoles of H2SO4. Volume = molars/concentration. Then it takes 14.96 mL of H2SO4.
The neutralization of h2so4aq using naohaq is represented by the following equation 2naohaq h2so4aq -- na2so4aq 2h2ol if 30.40 ml of 0.500 m naohaq is neutralized by 22.02 ml of?
To find the molarity of the sulfuric acid (H2SO4), we first calculate the number of moles of NaOH used: 30.40 mL * 0.500 mol/L = 15.20 mmol NaOH. Since the mole ratio between NaOH and H2SO4 is 2:1, 15.20 mmol of NaOH would neutralize 7.60 mmol of H2SO4. Now we can find the molarity of H2SO4 using its volume: 7.60 mmol / 22.02 mL = 0.345 M H2SO4.
What is the concentratuon of a hcl solution if 25.0 ml is required to neutralize 38.5 ml of a 0.500m solution of naoh?
To find the concentration of HCl, you can use the formula: moles of NaOH = moles of HCl. From the given information, you can calculate the moles of NaOH used to neutralize the acid. Then, use the volume and concentration of NaOH to determine the concentration of HCl.
What is the molarity of HNO3 if 20.0 ml of the solution is needed to exactly neutralize 10.0 ml of a 1.67 M NaOH solution?
The reaction between HNO3 and NaOH is a 1:1 molar ratio. This means that the moles of HNO3 required to neutralize the NaOH is the same as the moles of NaOH. Given that 20.0 ml of HNO3 is needed to neutralize 10.0 ml of a 1.67 M NaOH solution, the molarity of the HNO3 solution is twice the molarity of the NaOH solution, which is 3.34 M.
How many moles of NaOH react with 1 mole of H2SO4?
2 moles of NaOH will react with 1 mole of H2SO4 based on the balanced chemical equation: 2NaOH + H2SO4 -> Na2SO4 + 2H2O.
