A) 2
B) 1.77
C) 1.44
D) none of these
Correct Answer: B
Solution :
[b] \[Ti+{{O}_{2}}\xrightarrow{{}}T{{i}_{1.44}}O\] \[\frac{1.44}{48}mole\,\frac{x}{48(1.44)+16}mole\] \[\therefore \frac{1.44}{48}=\,\frac{1.44x}{48(1.44)+16}\] \[x=1.77g\]You need to login to perform this action.
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