A) 168 g
B) 84 g
C) 42 g
D) 21 g
Correct Answer: C
Solution :
\[3Fe+4{{H}_{2}}O\xrightarrow{{}}F{{e}_{3}}{{O}_{4}}+4{{H}_{2}}\] \[3\times 56g\,\,4\times 18g\] \[4\times 18g\,{{H}_{2}}O\equiv 3\times 56g\,Fe\] \[\therefore \] \[18g\,{{H}_{2}}O=\frac{3\times 56}{4}=42g\]You need to login to perform this action.
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