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A 24.803 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 79.733 g of water. A 10.560 g aliquot of this solution is then titrated with 0.1077 M HCl . It required 32.37 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.

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Answer:

2.37 (w/w)% of NH3 in the fertilizer

Explanation:

The HCl reacts with NH3 as follows:

HCl + NH3 ⇄ NH4Cl

To solve this question we must find the  moles of HCl used in the titration = Moles NH3. With its molar mass we can find mass of NH3 and using the dilutions we can find weight percent as follows:

Moles HCl = Moles NH3

32.37mL = 0.03237L * (0.1077mol/L) =

Mass NH3 in the dilution -Molar mass: 17.031g/mol-

0.003486moles NH3 * (17.031g/mol) = 0.05937g NH3

Mass NH3 in the sample:

0.05937g NH3 * (79.733g + 24.803g) / 10.560g =

0.588g NH3

Weight percent:

0.588g NH3 / 24.803g * 100 =

2.37 (w/w)% of NH3 in the fertilizer

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