Balanced equation.
H2SO4 + 2NaOH >> Na2SO4 + 2H2O
10 grams NaOH (1mol NaOH/39.998g )(1mol H2SO4/2mol NaOH )(98.086g H2SO4/1molH2SO4 ) = 12.26 grams of H2SO4
97.8 - 98.2 98 worked for me
Infinite because Sodium Chloride is neutral and will not neutralize sulfuric acid.
Balanced equation. 2NaOH + H2SO4 -> Na2SO4 + 2H2O 12.5 grams NaOH (1 mole NaOH/39.998 grams)(1 mole Na2SO4/2 mole NaOH)(142.05 grams/1 mole Na2SO4) = 22.2 grams sodium sulfate produced ===========================
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 )
98g of phosphoric acid require 120g of sodium hydroxide to produce sodium phosphate so 150g phosphoric acid will require = 120/98x150= 183.67g sodium hydroxide.
98g
262 - 266
262 - 266
97.8 - 98.2 98 worked for me
Infinite because Sodium Chloride is neutral and will not neutralize sulfuric acid.
1
It is 25 moles of Sodium Hydroxide (;
Balanced equation. 2NaOH + H2SO4 -> Na2SO4 + 2H2O 12.5 grams NaOH (1 mole NaOH/39.998 grams)(1 mole Na2SO4/2 mole NaOH)(142.05 grams/1 mole Na2SO4) = 22.2 grams sodium sulfate produced ===========================
155.2 g
The chemical equation is:2 NaOH + H2SO4 = Na2SO4 + 2 H2OMolar mass of sodium hydroxide is 39,9971 g; molar mass of sulfuric acid is 98,079 g.2 . 39,9971 g NaOH----------------------98,079 g H2SO4200 g NaOH------------------------xx = (200 x 98,079)/2 . 39,9971 = 245 g H2SO4So sulfuric acid is the limiting reagent.
the molar mass of sodium hydroxide is 40g/mol mike
how much sodium hydroxide in grams must be added to seawater to precipitate 86.9mg of magnesium present?