Satori F Edit Mew H. tory Bookmark Window Hop (d) If 46.2 mL of 0.108 M HCI solu

Question

Satori F Edit Mew H. tory Bookmark Window Hop (d) If 46.2 mL of 0.108 M HCI solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution Supporting Materials Periodic Table Fundamental 3. O-4 points BLB134 P082. (a) How many milliliters of 0.165 M HCI are needed to completely neutralize 25.0 mL of 0.101 M Ba(OH)2 solution? (b) How many milliliters of 0.180 M H2SO4 are needed to neutralize 0.150 g of NaOH? mL, (c) If 54.8 mL of is needed to precipitate all the sulfate io in a 762-mg sample of Nazso what is the molarity of the solution? Baci2 solution (d) if 28.7 solution is needed to neutralize a solution of Ca(oH)2, how many grams of Ca(oH)2 must be in the solution? mL of 0.168 M HCI

Explanation / Answer

Number (2) Issue (d). Answer: 42.6 mL of 0.108M hydrochloric acid containing 0.1821 grams of HCl . Through the stoichiometric ratio obtained from the neutralization reaction between HCl and KOH can be set that can neutralize HCl 36.5g KOH 62g , 0.1821 grams of HCl therefore can neutralize 0.3093 grams of KOH which are present in the aliquot initially achieved neutralize 46.2 mL of 0.108M HCL. Number (3) Problem (a). Answer: The volume of Hydrochloric Acid 0.165M needed to completely neutralize 25.0 mL of 0.101M Barium Hydroxide is 30.6 mL. (The data is obtained by performing the operations it so similiar to as explained in the Number (2) , problem (d)). Number (3) Problem (b). Answer: To neutralize 0.150 grams of Sodium Hydroxide from a 0.180M solution of sulfuric acid is needed to take a volume of 10.4 mL of the solution of sulfuric acid . For stoichiometric neutralization reaction is known to 98 grams of completely neutralize sulfuric acid 80 grams sodium hydroxide so 0150 grams of the substance will be neutralized by 0.18375 grams of sulfuric acid . With the amount of sulfuric acid (0.18375 grams) with molecular weight (98 g/mol) and the molar concentration known (0.180M) we can solve "volume" of the formula molarity and to obtain the value of the volume in LITERS (0.01041 L) is multiplied by 1000 to become "liters to mili-liters", which gives us the value of 10.4 mL of sulfuric acid 0.180M Number (3) Problem (d). Answer: 0.178 grams of Ca(OH)2 are present in solution. (The data is obtained by performing the operations it so similiar to as explained in the Number(2) , problem (d)).

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