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
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.
To understand the calculation, let's break it down step by step.
First, we have the initial reaction between NH4Cl (ammonium chloride) and NaOH (sodium hydroxide). The volume of NaOH used is 100 mL, and its concentration is 0.8N. To determine the amount of NaOH used, we multiply the volume (in liters) by the concentration (in moles per liter):
100 mL x 0.8N = 80.
This means that we used 80 moles of NaOH in the reaction.
Next, we have the neutralization of the excess NaOH with H2SO4 (sulfuric acid). The volume of H2SO4 used is 12.5 mL, and its concentration is 0.75N. Similarly, we multiply the volume (in liters) by the concentration (in moles per liter) to find the amount of H2SO4 used:
12.5 mL x 0.75N = 9.375.
This tells us that we used 9.375 moles of H2SO4 to neutralize the excess NaOH.
Now, to calculate the amount of NH4Cl initially taken, we subtract the moles of H2SO4 from the moles of NaOH:
80 - 9.375 = 70.625.
So, we had 70.625 moles of NH4Cl initially.
Finally, you mentioned the expression "70.625 x mew." I'm assuming "mew" is a typo, and you meant to write "molar mass (molecular weight)."
To find the amount of NH4Cl in grams, we can use its molar mass. The molar mass of NH4Cl is approximately 53.491 grams per mole. Therefore, we multiply the number of moles by the molar mass to obtain the mass in grams:
70.625 moles x 53.491 g/mol = 3776.28 grams, which is approximately 3776 grams of NH4Cl.
So, the amount of NH4Cl initially taken is approximately 3776 grams.