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The same molar amount, 0.100mol KOH. The reaction is in a 1:1 stoichiometric ratio: HCl + KOH --> KCl + H2O.
3 ions
the same amount.i.e.,hydrochloric acid reacts with caustic soda in the ratio 1/1.so if their concentrations are the same the amount of caustic soda required to neutralize hydrochloric acid will be same as that of amount of hydrochloric acid.
20 moles of NaOH needed to neutralize 20 moles of nitric acid
3.09
The same molar amount, 0.100mol KOH. The reaction is in a 1:1 stoichiometric ratio: HCl + KOH --> KCl + H2O.
The answer is o,13 g KOH.
3 ions
the same amount.i.e.,hydrochloric acid reacts with caustic soda in the ratio 1/1.so if their concentrations are the same the amount of caustic soda required to neutralize hydrochloric acid will be same as that of amount of hydrochloric acid.
20 moles of NaOH needed to neutralize 20 moles of nitric acid
3.09
it is defined as the number of miligrams of potassium hydroxide or sodium hydroxide required to sponify 1 gram of the fat or oil
0.1 moles
3
because alcohol is required for dissolving both KOH and lipid
Barium Hydroxide: Mass: 2.74g. Mr = 171. Moles = Mass (g)/Mr Therefore - 2.74/ 171 = 0.01602339181mols.
Alkalies can neutralize both concentrated and dilute acids, but dilute acids are 'more easily' neutralized (i.e. require a smaller amount of alkali for the same amount of acid). It is a simple chemical reaction, the amount of alkali required to neutralize an acid can be calculated if you know how they react and the strengths of the acid and alkali. E.g. if you use Potassium Hydroxide (KOH) to neutralize Hydrochloric acid (HCL) the reaction is: KOH + HCL --) KCL + H2O So one molecule of KOH neutralizes one molecule of HCL If you have 1 molar KOH, then : 10 ml of dilute HCL (0.1 molar strength) will be neutralized by 1 ml of KOH 10 ml of a strong HCL (10 molar strength) will be neutralized by 100 ml of KOH Hope that helps.