A) \[11.2\,mL\]
B) \[22.4\,mL\]
C) \[33.6\,mL\]
D) \[44.8\,mL\]
Correct Answer: D
Solution :
\[PbS+4{{H}_{2}}{{O}_{2}}\to PbS{{O}_{4}}+\underset{Water}{\mathop{4{{H}_{2}}O}}\,\] \[\underset{=136\,g}{\mathop{\underset{4(2+32)}{\mathop{4{{H}_{2}}{{O}_{2}}}}\,}}\,\to \underset{at\,NTP.}{\mathop{\underset{2\times 22.4\,litre}{\mathop{4{{H}_{2}}O+2{{O}_{2}}}}\,}}\,\] From the above equation we can derive that 1 mole of \[PbS\]reacts with \[0.01\times 4\] i.e. 0.04 moles of \[{{H}_{2}}{{O}_{2}}\] Thus, volume of \[{{H}_{2}}{{O}_{2}}\] needed \[=\frac{0.04\times 11.2}{10}=0.0448L\]or \[44.8\text{ }mL.\]
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