| \[{{H}_{2}}{{O}_{2}}+2KI\xrightarrow{40%yield}{{I}_{2}}+2KOH\] |
| \[{{H}_{2}}{{O}_{2}}+2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\xrightarrow{50%yield}\]\[{{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+3{{O}_{2}}+4{{H}_{2}}O\] |
| 150 mL of \[{{H}_{2}}{{O}_{2}}\] sample was divided into two parts. First part was treated with KI and formed KOH required 200 mL of M/2 \[{{H}_{2}}S{{O}_{4}}\] for neutralization. Other part was treated with \[KMn{{O}_{4}}\] yielding 6.74 litre of \[{{O}_{2}}\] at STP. Using % yield indicated find volume strength of \[{{H}_{2}}{{O}_{2}}\]sample used. |
| \[{{H}_{2}}{{O}_{2}}+2KI\xrightarrow{40%yield}{{I}_{2}}+2KOH\] |
| \[{{H}_{2}}{{O}_{2}}+2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\xrightarrow{50%yield}\]\[{{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+3{{O}_{2}}+4{{H}_{2}}O\] |
A) 5.04
B) 10.08
C) 3.36
D) 33.6
Correct Answer: D
Solution :
| Mole of \[{{H}_{2}}S{{O}_{4}}=0.1,\] mole of KOH = 0.2 Mole of \[{{H}_{2}}S{{O}_{4}}\] used in first reaction \[=\frac{0.2}{2}\,\,\times \,\,\frac{1}{0.4}=0.25\] |
| Mole of produced \[{{O}_{2}}=\frac{6.74}{22.4}=0.3\] |
| Mole of used in second reaction \[=\frac{0.3}{3\,\,\times \,\,0.5}=0.2\] |
| Total mole of consumed \[{{H}_{2}}{{O}_{2}}=0.45\] |
| Molarity of \[{{H}_{2}}{{O}_{2}}=\frac{0.45}{0.15}=3M\] |
| Volume strength \[112\,\,\times \,\,3=33.6\] |
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