A) 83.4 g
B) 16.7 g
C) 1.67 g
D) 8.3 g
Correct Answer: A
Solution :
\[\underset{\begin{smallmatrix} 2\times 158 \\ =316g \end{smallmatrix}}{\mathop{2KMn{{O}_{4}}}}\,+\underset{10\times 278=2780\,g}{\mathop{10FeS{{O}_{4}}.7{{H}_{2}}O}}\,+8{{H}_{2}}S{{O}_{4}}\] \[\xrightarrow{{}}5F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+{{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}\] \[+78{{H}_{2}}O\] 316 g\[KMn{{O}_{4}},\]oxidises\[=2780\,g\,FeS{{O}_{4}}.7{{H}_{2}}O\] 9.48 g\[KMn{{O}_{4}}\]will oxidise\[=\frac{2780}{316}\times 9.48\] \[=83.4\,g\,FeS{{O}_{4}}.7{{H}_{2}}O\]
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