ml x N = number of milliequivalents 80 KOH added initially. The KOH reacted with the NH3 produced from the reaction of NH4Cl with the KOH). After the reaction the excess KOH was titrated with 12.5 mL of 0.75 N acid.
12.5 x 0.75 N = 9.375 milliequivalents acid needed to neutralize the EXCESS KOH. Therefore, the NH4Cl there initially is
80 m.e. initially
-9.375 m.e. not used by the NH4Cl
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70.625 milliequialents NH4Cl there initially.
How many grams NH4Cl is this? That's
#milliequivalents x milliequivalent weight = grams. (That's the same thing as grams = mols x molar mass except we are dealing with equivalents in the normal system instead of moles in the molar system.)
70.625 x 0.053491 = 3.80 grams NH4Cl there initially.
A definite amount of NH4cl was boiled with 100ml of 0.8N naoh for complete reaction . After the reaction mixture containing excess of naoh was neutralized with 12.5ml of 0.75N NH2so4. Calculate the amount of NH4cl taken.
I want you to explain about
100ml x 0.8N = 80
12.5ml x 0.75 =9.375
80 - 9.375 = 70.625
70.625 x mew This part I do not understand
NH4CL = 53.491
1 answer
1 answer