A) \[1.44\]
B) \[1.77\]
C) \[1.33\]
D) \[1.55\]
Correct Answer: B
Solution :
In the reaction \[Ti+{{O}_{2}}\xrightarrow{{}}T{{i}_{1.44}}O\] on applying POAC for Ti-atom, \[\therefore \]Number of moles of Titanium\[=1.44\times \]Number of moles of \[T{{i}_{1.44}}O\] \[\frac{1.44}{48}=\frac{1.44x}{48\times 1.44+16}\] \[x=1.77g\]You need to login to perform this action.
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