2 M HCl means that the concentration is 2 mols per litre. Many starting chemists have a hard time remembering that, probably because it's a contraction that really isn't required. But there you have it. If someone says "2 molar", that's 2 mols per litre.
Anyhow, half a litre is obviously 1 mol HCl, in this case. Then we just need to know that one HCl molecule is neutralized by one OH- ion. (We can easily see this since HCl exists as H+ and Cl-, and the H+ ion content is how we commonly gauge acidity of solution. Then we see that H+ and OH- can form H2O, neutralising the solution).
Thus, the answer is 1 mol of OH-.
20 moles of NaOH needed to neutralize 20 moles of nitric acid
3 moles KOH to neutralize 3 moles HNO3 : 3 mol OH- will react with 3 mol H+
Molarity = moles of solute/liters of solution or, for our purposes moles of solute = liters of solution * Molarity moles of AgNO3 = 0,50 liters * 4.0 M = 2.0 moles of AgNO3 needed --------------------------------------
Balanced equation. NaOH + HCl -> NaCl + H2O all one to one. find moles HCl. 11 grams HCl (1 mole HCl/36.458 grams ) = 0.3017 moles HCl Moles HCl same as moles NaOH Molarity = moles of solute/Liters of solution 1.06 M NaOH = 0.3017 moles NaOH/liters of solution = 0.2846 Liters this is equal to..... 285 milliliters of NaOH needed
The answer is 7,66 L.
20 moles of NaOH needed to neutralize 20 moles of nitric acid
3 moles KOH to neutralize 3 moles HNO3 : 3 mol OH- will react with 3 mol H+
Molarity = moles of solute/liters of solution or, for our purposes moles of solute = liters of solution * Molarity moles of AgNO3 = 0,50 liters * 4.0 M = 2.0 moles of AgNO3 needed --------------------------------------
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.
0.125 Molar solution! Molarity = moles of solute/Liters of solution Algebraically manipulated, Moles of copper sulfate = 2.50 Liters * 0.125 M = 0.313 moles copper sulfate needed ===========================
Balanced equation. NaOH + HCl -> NaCl + H2O all one to one. find moles HCl. 11 grams HCl (1 mole HCl/36.458 grams ) = 0.3017 moles HCl Moles HCl same as moles NaOH Molarity = moles of solute/Liters of solution 1.06 M NaOH = 0.3017 moles NaOH/liters of solution = 0.2846 Liters this is equal to..... 285 milliliters of NaOH needed
Several part problem. Get molarity of NaHCO3. (150 ml)( M NaHCO3) = (150 ml)(0.44 M HCl) = 0.44 M NaHCO3 --------------------------- get moles NaHCO3 ( 150 ml = 0.150 Liters ) 0.44 M NaHCO3 = moles NaHCO3/0.150 Liters = 0.066 moles NaHCO3 ---------------------------------------get grams 0.066 moles NaHCO3 (84.008 grams/1 mole NaHCO3) = 5.54 grams NaHCO3 needed ---------------------------------------------answer
The answer is 7,66 L.
Balanced equation first, last and always! Na2CO3 + 2HCl - > 2NaCl + CO2 + H2O 2.5 g Na2CO3 ( 1 mole Na2CO3/105.99 g)(2 mole HCl/1 mole Na2CO3) 0.04717 moles HCl -------------------------Now, Molarity = moles of solute/liters of solution or, for our purposes liters of solution = moles of solute/Molarity Liters HCl = 0.04717 moles HCl/0.60 M HCl = 0.0786 liters (1000 milliliters/1 liter) = 78. 6 milliliters HCl solution needed
You need 0,5 L.
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)
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