100ml / 1000 = 0.1mol
*1M = divide by 1000
moles = mass / molecular mass
mass = mole x molecular mass of nacl
mass = 0.1 x 58.5
mass = 5.85
answer is 5.85 grams
To completely neutralize 100ml of 1M H2SO4, you would need an equal number of moles of NaCl. H2SO4 is a diprotic acid, so it will require 2 moles of NaCl to neutralize 1 mole of H2SO4. Therefore, you would need 2 moles of NaCl for every mole of H2SO4. With a 1M solution of H2SO4 in 100ml, you have 0.1 moles of H2SO4. Therefore, you would need 0.2 moles of NaCl. The molar mass of NaCl is approximately 58.44g/mol, so you would need approximately 11.7 grams of NaCl to completely neutralize the 1M H2SO4 solution.
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!
To calculate the amount of limestone needed, first determine the moles of H2SO4 and HNO3 in the lake. Then, use the stoichiometry of the neutralization reaction between limestone (CaCO3) and the acids to calculate the moles of limestone required to neutralize the acids. Finally, convert the moles of limestone to kilograms using the molar mass of CaCO3.
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.
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.
NaCl doesn't neutralize sulfuric acid.
To completely neutralize 100ml of 1M H2SO4, you would need an equal number of moles of NaCl. H2SO4 is a diprotic acid, so it will require 2 moles of NaCl to neutralize 1 mole of H2SO4. Therefore, you would need 2 moles of NaCl for every mole of H2SO4. With a 1M solution of H2SO4 in 100ml, you have 0.1 moles of H2SO4. Therefore, you would need 0.2 moles of NaCl. The molar mass of NaCl is approximately 58.44g/mol, so you would need approximately 11.7 grams of NaCl to completely neutralize the 1M H2SO4 solution.
Strong base is added to neutralize the strong acid (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!
To calculate the amount of limestone needed, first determine the moles of H2SO4 and HNO3 in the lake. Then, use the stoichiometry of the neutralization reaction between limestone (CaCO3) and the acids to calculate the moles of limestone required to neutralize the acids. Finally, convert the moles of limestone to kilograms using the molar mass of CaCO3.
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.
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.
Grams NaOH?? Balanced equation. 2NaOH + H2SO4 --> Na2SO4 + 2H2O 4.9 grams H2SO4 (1 mole H2SO4/98.086 grams)(2 mole NaOH/1 mole H2SO4)(39.998 grams/1 mole NaOH) = 4.0 grams NaOH needed =================
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.)
To find the number of moles of H2SO4 in 100 mL of 12M H2SO4 solution, you first need to convert the volume to liters, which is 0.1 L. Then, you multiply the volume in liters by the molarity to get the moles of H2SO4. So, 0.1 L * 12 mol/L = 1.2 moles of H2SO4.
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.
The balanced chemical equation for the reaction is: 2 KOH + H2SO4 -> K2SO4 + 2 H2O From the equation, it can be seen that 2 moles of KOH react with 1 mole of H2SO4. Calculate the moles of H2SO4 (2.70 g / molar mass of H2SO4) and then use the mole ratio to find the moles of KOH required. Finally, convert the moles of KOH to mass (moles of KOH x molar mass of KOH) to get the required mass of KOH.