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Please answer the following question with explanation
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Asked by deepakudgiri29 | 25 Dec, 2018, 05:19: PM
answered-by-expert Expert Answer
The reaction is as;
 
(NH4)2 SO4 + 2NaOH → 2NH3 + Na2SO4 + 2H2O
 
2NaOH +  H2SO4 →  Na2SO4 + 2H2O
 
5 ml 0.2 N H2SO4 solution used for titration,
So,
 
0.2 N H2SO4 = 0.1 M H2SO4 
 
 
No. space of space moles space of space straight H subscript 2 SO subscript 4 space equals space 0.1 cross times 0.005 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 5 cross times 10 to the power of negative 3 space end exponent space moles From space reaction space we space know space that comma 2 space moles space of space NaOH space requires space 1 space mole space of space straight H subscript 2 SO subscript 4 therefore space space 5 cross times 10 to the power of negative 3 space end exponent space moles space of space straight H subscript 2 SO subscript 4 space will space require space 1 cross times 10 to the power of negative 2 space end exponent space moles space of space NaOH Now space from space 250 space ml space 25 space ml space solution space was space taken space for space analysis. So comma no. space of space moles space in space 25 space ml space is space 1 cross times 10 to the power of negative 2 space end exponent cross times 10 space equals space 0.01 space moles The space no. space of space moles space of space NaOH space remained space after space reaction space with space open parentheses NH subscript 4 close parentheses subscript 2 SO subscript 4 Now comma space no. space of space moles space NaOH space in space 100 space ml space equals space 0.2 space cross times 0.1 space equals space space 0.02 space moles therefore space moles space of space NaOH space reacted space with space open parentheses NH subscript 4 close parentheses subscript 2 SO subscript 4 space equals space 0.2 space minus space 0.01 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.01 space mole From space reaction comma space we space know space that comma 2 space moles space of space NaOH space reacts space with space 1 space mole space of space open parentheses NH subscript 4 close parentheses subscript 2 SO subscript 4 space so comma space 0.01 space mole space of space NaOH space reacts space with space 0.005 space moles space of space open parentheses NH subscript 4 close parentheses subscript 2 SO subscript 4. We space have comma no. space of space moles space equals fraction numerator Mass over denominator Mol. space mass end fraction Mass space of space open parentheses NH subscript 4 close parentheses subscript 2 SO subscript 4 space equals space 0.005 space cross times 132.14 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.66 space straight g The space initial space weight space given space is space 0.80 space straight g percent sign space Purity space equals fraction numerator 0.66 over denominator 0.8 end fraction cross times 100 space space space space space space space space space space space space space space space equals space 82.5 percent sign
 
The percentage purity of the (NH4)2 SO4 samples is 82.5%.
Answered by Varsha | 26 Dec, 2018, 04:37: PM

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